(C) - KSUfac.ksu.edu.sa/sites/default/files/mdhkr_qdym_lmqrr_104...c) ertices: V (0 b), V 2 (0; b) 2...
Transcript of (C) - KSUfac.ksu.edu.sa/sites/default/files/mdhkr_qdym_lmqrr_104...c) ertices: V (0 b), V 2 (0; b) 2...
![Page 1: (C) - KSUfac.ksu.edu.sa/sites/default/files/mdhkr_qdym_lmqrr_104...c) ertices: V (0 b), V 2 (0; b) 2 b. 2 a ts: W 1 (a; 0), W 2 (a; 0). 7 aph. (a) 9 x 2 25 y 2 225 (b) 16 x 2 9 y 2](https://reader035.fdocuments.us/reader035/viewer/2022062415/5fdb545c9fe12d2da100204e/html5/thumbnails/1.jpg)
![Page 2: (C) - KSUfac.ksu.edu.sa/sites/default/files/mdhkr_qdym_lmqrr_104...c) ertices: V (0 b), V 2 (0; b) 2 b. 2 a ts: W 1 (a; 0), W 2 (a; 0). 7 aph. (a) 9 x 2 25 y 2 225 (b) 16 x 2 9 y 2](https://reader035.fdocuments.us/reader035/viewer/2022062415/5fdb545c9fe12d2da100204e/html5/thumbnails/2.jpg)
![Page 3: (C) - KSUfac.ksu.edu.sa/sites/default/files/mdhkr_qdym_lmqrr_104...c) ertices: V (0 b), V 2 (0; b) 2 b. 2 a ts: W 1 (a; 0), W 2 (a; 0). 7 aph. (a) 9 x 2 25 y 2 225 (b) 16 x 2 9 y 2](https://reader035.fdocuments.us/reader035/viewer/2022062415/5fdb545c9fe12d2da100204e/html5/thumbnails/3.jpg)
M-1
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R1:
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(1)
Para
bola
Defi
nit
ion
1A
para
bola
isth
ese
tof
all
poin
tsin
the
pla
ne
equ
idis
tan
tfr
om
afi
xed
poin
tF
(call
edth
efo
cus)
an
da
fixe
dli
neD
(call
edth
edir
ectr
ix)
inth
esa
me
pla
ne.
(1)
Th
evert
ex
of
the
para
bola
isth
eori
gin
(0,0
):
(A)x2
=4ay,
wh
erea>
0•
Ver
tex:V
(0,0
)•
Itop
ens
upw
ard
s•
Its
axis
:th
ey-a
xis
•F
ocu
s:isF
(0,a
)•
Dir
ectr
ix:y
=−a
(B)x2
=−
4ay,
wh
erea>
0•
Ver
tex:V
(0,0
)•
Itop
ens
dow
nw
ard
s•
Its
axis
:y-a
xis
•F
ocu
s:F
(0,−a)
•D
irec
trix
:y
=a
(C)y2
=4ax
,w
her
ea>
0•
Ver
tex:V
(0,0
)•
Itop
ens
toth
eri
ght
•It
sax
is:
the
x-a
xis
•F
ocu
sisF
(a,0
)•
Dir
ectr
ix:x
=−a
(D)y2
=−
4ax
,w
her
ea>
0•
Ver
tex:V
(0,0
)•
Itop
ens
toth
ele
ft•
Its
axis
:x-a
xis
•F
ocu
s:isF
(−a,0
)•
Dir
ectr
ix:x
=a
Exam
ple
1F
ind
the
focu
san
dth
edir
ectr
ixof
the
para
bolax2
=4y
,an
dsk
etch
its
graph.
Exam
ple
2F
ind
the
focu
san
dth
edir
ectr
ixof
the
para
bolay2
=−
8x,
an
dsk
etch
its
graph.
Exam
ple
3F
ind
the
equ
ati
on
of
the
para
bola
wit
hfo
cus
(3,0
)an
ddi-
rect
rixx
=−
3.T
hen
,sk
etch
the
graph.
![Page 4: (C) - KSUfac.ksu.edu.sa/sites/default/files/mdhkr_qdym_lmqrr_104...c) ertices: V (0 b), V 2 (0; b) 2 b. 2 a ts: W 1 (a; 0), W 2 (a; 0). 7 aph. (a) 9 x 2 25 y 2 225 (b) 16 x 2 9 y 2](https://reader035.fdocuments.us/reader035/viewer/2022062415/5fdb545c9fe12d2da100204e/html5/thumbnails/4.jpg)
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(2)
Th
egen
era
lfo
rmu
laof
ap
ara
bola
:
(A)
(x−h
)2=
4a(y−k),
wh
erea>
0•
Ver
tex:V
(h,k
)•
Itop
ens
upw
ard
s•
Its
axis
:p
aral
lel
toth
ey-a
xis
•F
ocu
s:F
(h,k
+a)
•D
irec
trix
:y
=k−a
(B)
(x−h
)2=−
4a(y−k),
wh
ere
a>
0 •V
erte
x:V
(h,k
)•
Its
axis
:p
ara
llel
toth
ey-a
xis
•It
open
sd
ownw
ard
s•
Focu
s:F
(h,k−a)
•D
irec
trix
:y
=k
+a
(C)
(y−k)2
=4a
(x−h
),w
her
ea>
0•
Ver
tex:V
(h,k
)•
Its
axis
:p
aral
lel
toth
ex-a
xis
•It
open
sto
the
righ
t•
Focu
s:F
(h+a,k
)•
Dir
ectr
ix:x
=h−a
(D)
(y−k)2
=−
4a(x−h
),w
her
ea>
0•
Ver
tex:V
(h,k
)•
Itop
ens
toth
ele
ft•
Focu
s:F
(h−a,k
)•
Dir
ectr
ix:x
=h
+a
•It
saxis
:p
ara
llel
toth
ex-a
xis
Exam
ple
4F
ind
the
focu
san
dth
edir
ectr
ixof
the
para
bola
(x+
1)2
=−
4(y−
1)
,an
dsk
etch
its
graph.
Exam
ple
5F
ind
the
equ
ati
on
of
the
para
bola
wit
hve
rtex
(2,1
)an
dfo
cusF
(2,3
).T
hen
,sk
etch
the
graph.
Exam
ple
6F
ind
the
focu
san
dth
edir
ectr
ixof
the
para
bola
2y2−
4y+
8x
+10
=0
,an
dsk
etch
its
graph.
![Page 5: (C) - KSUfac.ksu.edu.sa/sites/default/files/mdhkr_qdym_lmqrr_104...c) ertices: V (0 b), V 2 (0; b) 2 b. 2 a ts: W 1 (a; 0), W 2 (a; 0). 7 aph. (a) 9 x 2 25 y 2 225 (b) 16 x 2 9 y 2](https://reader035.fdocuments.us/reader035/viewer/2022062415/5fdb545c9fe12d2da100204e/html5/thumbnails/5.jpg)
M-1
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NE
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LM
AT
HE
MA
TIC
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CH
AP
TE
R1:
CO
NIC
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SD
ate
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ay:
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(2)
Ell
ipse
Defi
nit
ion
2A
nel
lipse
isth
ese
tof
all
poin
tsin
the
pla
ne
for
whic
hth
esu
mof
the
dis
tan
ces
totw
ofi
xed
poin
tsis
con
stan
t.
(1)
Th
ecente
rof
the
ell
ipse
isth
eori
gin
(0,0
):
(A)x2
a2
+y2
b2
=1
wh
ere
a>b
c=√ a2
−b2
•C
ente
r:P
(0,0
)•
Foci
:F1(−c,
0),F2(c,0
).•
Ver
tice
s:V1(−a,0
),V2(a,0
).•
Ma
jor
axis
:th
ex-a
xis
,le
ngt
his
2a.
•M
inor
axis
:y-a
xis
,le
ngt
his
2b.
•M
inor
axis
end
poi
nts
:W
1(0,b
),W
2(0,−b)
.
(B)x2
a2
+y2
b2
=1
wh
ere
a<b
c=√ b2
−a2
•C
ente
r:P
(0,0
)•
Foci
:F1(0,c
),F2(0,−c)
•V
erti
ces:V1(0,b
),V2(0,−b)
•M
ajo
rax
is:
y-a
xis
,le
ngt
his
2b.
•M
inor
axis
:x-a
xis
,le
ngt
his
2a.
•M
inor
axis
end
poi
nts
:W
1(−a,0
),W
2(a,0
).
Exam
ple
7Id
enti
fyth
efe
atu
res
of
the
elli
pse
an
dsk
etch
its
graph.
(a)
9x2
+25y2
=22
5(b
)16x2
+9y2
=14
4
Exam
ple
8F
ind
an
equ
ati
on
of
an
elli
pse
ifth
ece
nte
ris
at
the
ori
gin
an
d
(a)
Majo
raxi
son
x-axi
sM
ajo
raxi
sle
ngt
h=
14
Min
or
axi
sle
ngt
h=
10
(b)
Majo
raxi
son
y-axi
sM
inor
axi
sle
ngt
h=
14
Dis
tan
ceof
foci
fro
mce
nte
r=
10√
2
![Page 6: (C) - KSUfac.ksu.edu.sa/sites/default/files/mdhkr_qdym_lmqrr_104...c) ertices: V (0 b), V 2 (0; b) 2 b. 2 a ts: W 1 (a; 0), W 2 (a; 0). 7 aph. (a) 9 x 2 25 y 2 225 (b) 16 x 2 9 y 2](https://reader035.fdocuments.us/reader035/viewer/2022062415/5fdb545c9fe12d2da100204e/html5/thumbnails/6.jpg)
M-1
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NE
RA
LM
AT
HE
MA
TIC
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AP
TE
R1:
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SD
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==
(2)
Th
egen
era
lfo
rmu
laof
the
ell
ipse
:
(A)
(x−h)2
a2
+(y−k)2
b2
=1
wh
ere
a>b
,c
=√a2−b2
•C
ente
r:P
(h,k
)•
Foci
:F1(h−c,k),F2(h
+c,k)
•V
erti
ces:V1(h−a,k
),V2(h
+a,k
)•
Ma
jor
axis
:p
aral
lel
toth
ex-a
xis
,le
ngt
his
2a
•M
inor
axis
:p
aral
lel
toth
ey-a
xis
,le
ngt
his
2b•
Min
orax
isen
dp
oints
:W
1(h,k
+b)
,W
2(h,k−b)
(B)
(x−h)2
a2
+(y−k)2
b2
=1
wh
ere
a<b
,c
=√b2−a2
•C
ente
r:P
(h,k
)•
Foci
:F1(h,k
+c)
,F2(h,k−c)
•V
erti
ces:V1(h,k
+b)
,V2(h,k−b)
•M
ajo
rax
is:
par
alle
lto
the
y-a
xis
,le
ngt
his
2b
•M
inor
axis
:p
aral
lel
toth
ex-a
xis
,le
ngt
his
2a•
Min
orax
isen
dp
oints
:W
1(h−a,k
)an
dW
2(h
+a,k
)
Exam
ple
9F
ind
the
equ
ati
on
of
the
elli
pse
wit
hfo
ciat
(−3,1
),
(5,1
),
an
don
eof
its
vert
ice
is(7,1
),
an
dsk
etch
its
graph.
Exam
ple
10
Fin
dth
eeq
uati
on
of
the
elli
pse
wit
hfo
ciat
(2,5
),
(2,−
3)
,an
dth
ele
ngt
hof
its
min
or
axi
seq
uals
6,
an
dsk
etch
its
graph.
Exam
ple
11
Iden
tify
the
featu
res
of
the
elli
pse
4x2+
2y2−
8x−
8y−
20
=0,
an
dsk
etch
its
graph.
![Page 7: (C) - KSUfac.ksu.edu.sa/sites/default/files/mdhkr_qdym_lmqrr_104...c) ertices: V (0 b), V 2 (0; b) 2 b. 2 a ts: W 1 (a; 0), W 2 (a; 0). 7 aph. (a) 9 x 2 25 y 2 225 (b) 16 x 2 9 y 2](https://reader035.fdocuments.us/reader035/viewer/2022062415/5fdb545c9fe12d2da100204e/html5/thumbnails/7.jpg)
M-1
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NE
RA
LM
AT
HE
MA
TIC
S2
CH
AP
TE
R1:
CO
NIC
SE
CT
ION
SD
ate
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ay:
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(3)
Hyp
erb
ola
Defi
nit
ion
3A
hyp
erbo
lais
the
set
of
all
poin
tsin
the
pla
ne
for
whic
hth
ediff
eren
ceof
the
dis
tan
ces
betw
een
two
fixe
dpo
ints
isco
nst
an
t.
(1)
Th
ecente
rof
the
hyp
erb
ola
isth
eori
gin
(0,0
):
c=√ a2
+b2
Th
eli
ne
segm
ent
bet
wee
nV1
andV2
isth
etr
an
sver
seaxis
.
(A)x2
a2−
y2
b2
=1
•C
ente
r:P
(0,0
)•
Foci
:F1(−c,
0),F2(c,0
)•
Ver
tice
s:V1(−a,0
),V2(a,0
)•
Tra
nsv
erse
axis
:x-a
xis
,le
ngt
his
2a.
•A
sym
pto
tes:y
=±b ax
.
(B)y2
b2−
x2
a2
=1
•C
ente
r:P
(0,0
)•
Foci
:F1(0,c
),F2(0,−c)
•V
erti
ces:V1(0,b
),V2(0,−b)
•T
ran
sver
seax
is:
y-a
xis
,le
ngt
his
2b•
Asy
mp
tote
s:y
=±b ax
Exam
ple
12
Iden
tify
the
featu
res
of
the
hyp
erbo
laan
dsk
etch
its
graph.
(a)
4x2−
16y2
=64
(b)
4y2−
9x2
=36
Exam
ple
13
Fin
dan
equ
ati
on
of
ahyp
erbo
laif
its
vert
ices
areV1(3,0
)an
dV2(−
3,0)
an
don
eof
its
foci
(4,0
).
![Page 8: (C) - KSUfac.ksu.edu.sa/sites/default/files/mdhkr_qdym_lmqrr_104...c) ertices: V (0 b), V 2 (0; b) 2 b. 2 a ts: W 1 (a; 0), W 2 (a; 0). 7 aph. (a) 9 x 2 25 y 2 225 (b) 16 x 2 9 y 2](https://reader035.fdocuments.us/reader035/viewer/2022062415/5fdb545c9fe12d2da100204e/html5/thumbnails/8.jpg)
M-1
04
GE
NE
RA
LM
AT
HE
MA
TIC
S2
CH
AP
TE
R1:
CO
NIC
SE
CT
ION
SD
ate
:/
/D
ay:
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(2)
Th
egen
era
lfo
rmu
laof
the
hyp
erb
ola
:
c=√ a2
+b2
(A)
(x−h)2
a2−
(y−k)2
b2
=1
•C
ente
r:P
(h,k
)•
Foci
:F1(h−c,k),F2(h
+c,k)
•V
erti
ces:V1(h−a,k
),V2(h
+a,k
)•
Tra
nsv
erse
axis
:p
aral
lel
tox-a
xis
,le
ngt
his
2a
•A
sym
pto
tes:
(y−k)
=±b a(x−h
)
(B)
(y−k)2
b2−
(x−h)2
a2
=1
•C
ente
r:P
(h,k
)•
Foci
:F1(h,k
+c)
,F2(h,k−c)
•V
erti
ces:V1(h,k
+b)
,V2(h,k−b)
•T
ran
sver
seax
is:
par
alle
lto
y-a
xis
,le
ngt
his
2b
•A
sym
pto
tes:
(y−k)
=±b a(x−h
)
Exam
ple
14
Fin
dth
eeq
uati
on
of
the
hyp
erbo
law
ith
foci
at
(−2,2
),
(6,2
)an
don
eof
its
vert
ices
is(5,2
),
an
dsk
etch
its
graph.
Exam
ple
15
Fin
dth
eeq
uati
on
of
the
hyp
erbo
law
ith
foci
at
(−1,−
6)
,(−
1,4
)an
dth
ele
ngt
hof
its
tran
sver
seaxi
sis
8,
an
dsk
etch
its
graph.
Exam
ple
16
Iden
tify
the
featu
res
of
the
hyp
erbo
la2y
2−
4x2−
4y−
8x−
34
=0,
an
dsk
etch
its
graph.
![Page 9: (C) - KSUfac.ksu.edu.sa/sites/default/files/mdhkr_qdym_lmqrr_104...c) ertices: V (0 b), V 2 (0; b) 2 b. 2 a ts: W 1 (a; 0), W 2 (a; 0). 7 aph. (a) 9 x 2 25 y 2 225 (b) 16 x 2 9 y 2](https://reader035.fdocuments.us/reader035/viewer/2022062415/5fdb545c9fe12d2da100204e/html5/thumbnails/9.jpg)
M-1
04
GE
NE
RA
LM
AT
HE
MA
TIC
S2
CH
AP
TE
R1:
CO
NIC
SE
CT
ION
SD
ate
:/
/D
ay:
Page:
07
==
==
==
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==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
Hom
ew
ork
(1):
1.F
ind
the
focu
san
dth
edir
ectr
ixof
the
par
ab
ola
(x−
1)2
=8(y
+1),
and
sket
chit
sgr
aph
.
2.F
ind
the
equ
atio
nof
the
par
abol
aw
ith
vert
ex(−
4,2)
an
dfo
cus
(−7 2,2
).T
hen
,sk
etch
the
grap
h.
3.F
ind
the
equ
atio
nof
the
par
abol
aw
ith
focu
s(3,6
)an
dd
irec
trix
y=
2.T
hen
,sk
etch
the
grap
h.
4.F
ind
aneq
uat
ion
ofan
elli
pse
ifth
ece
nte
ris
at
the
ori
gin
an
dm
ajo
rax
isis
onx-a
xis
and
its
len
gth
equ
als
8an
dm
inor
axis
len
gth
equ
als
6.
5.F
ind
the
equ
atio
nof
the
elli
pse
wit
hfo
ciat
(10,−
2),
(4,−
2),
an
don
eof
its
vert
ices
is(1
2,−
2),
then
sket
chit
sgr
aph
.
6.
Fin
dan
equ
ati
on
of
ahyp
erb
ola
ifit
sve
rtic
esare
(0,−
2)
an
d(0,2
)an
don
eof
its
foci
(0,√
13).
7.
Fin
dth
eeq
uati
on
of
the
hyp
erb
ola
wit
hfo
ciat
(4,−
2),
(10,−
2)
and
on
eof
its
vert
ices
is(8,−
2),
an
dsk
etch
its
gra
ph
.
8.
Fin
dth
eel
emen
tsof
the
con
icse
ctio
nx2
+5y2
+6x−
40y
+84
=0,
an
dsk
etch
its
gra
ph
.
9.
Fin
dth
eel
emen
tsof
the
con
icse
ctio
nx2
+2y
+2x
=2,
an
dsk
etch
its
gra
ph
.
10.
Fin
dth
eel
emen
tsof
the
con
icse
ctio
ny2−
5x2
+6y−
40x−
76
=0,
an
dsk
etch
its
gra
ph
.
![Page 10: (C) - KSUfac.ksu.edu.sa/sites/default/files/mdhkr_qdym_lmqrr_104...c) ertices: V (0 b), V 2 (0; b) 2 b. 2 a ts: W 1 (a; 0), W 2 (a; 0). 7 aph. (a) 9 x 2 25 y 2 225 (b) 16 x 2 9 y 2](https://reader035.fdocuments.us/reader035/viewer/2022062415/5fdb545c9fe12d2da100204e/html5/thumbnails/10.jpg)
M-1
04
GE
NE
RA
LM
AT
HE
MA
TIC
S2
CH
AP
TE
R2:
Matr
ices
an
dD
ete
rmin
ants
Date
:/
/D
ay:
Page:
08
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
(1)
Matr
ices
Defi
nit
ion
4A
matr
ixA
of
ord
erm×n
isa
set
of
real
nu
mbe
rsarr
an
ged
ina
rect
an
gula
rarr
ay
ofm
row
san
dn
colu
mn
s.
A=
a11
a12···
a1n
a21
a22···
a2n
. . .. . .
. ..
. . .am
1am
2···amn
N
ote
s:
1.aij
rep
rese
nts
the
elem
ent
ofth
em
atri
xA
that
lies
inro
wi
an
dco
lum
nj.
2.T
he
mat
rixA
can
also
be
wri
tten
asA
=(aij
) m×n.
3.If
the
nu
mb
erof
row
seq
ual
sth
enu
mb
erof
colu
mn
s(m
=n
),th
enA
isca
lled
asq
uar
em
atri
xof
ord
ern
.4.
Ina
squ
are
mat
rixA
=(aij
),
the
set
ofel
emen
tsof
the
form
aii
isca
lled
the
dia
gon
alof
the
mat
rix.
Exam
ple
17
Fin
dth
eord
erof
each
matr
ix,
then
fin
dth
eel
emen
ts:
1.
A=
( 2−
41
0
) ,a11
an
da22
2.
A=
( 13
52
10
) ,a12,a21
an
da23
Exam
ple
18
Fin
dth
edia
gon
al
of
the
squ
are
matr
ix.
1.
A=
( 2−
41
0
)
2.
A=
2−
73
10
9−
16
8
Sp
ecia
lty
pes
of
matr
ices:
1.
Row
vecto
r:
Aro
wve
ctor
of
ord
ern
isa
matr
ixof
ord
er1×n
,an
dit
isw
ritt
enas
(a1a2...an)
Exam
ple
:(2
70−
19)
isa
row
vect
or
of
ord
er5.
2.
Colu
mn
vecto
r:
Aco
lum
nve
ctor
of
ord
ern
isa
matr
ixof
ord
er
n×
1,
an
dit
isw
ritt
enas
A=
a1
a2 . . . an
E
xam
ple
:
1 7 3
isa
colu
mn
vect
or
of
ord
er3.
3.
Nu
llm
atr
ix:
Th
em
atr
ix(aij
) m×n
of
ord
erm×n
isca
lled
anu
llm
atr
ixifaij
=0
for
alli
an
dj,
an
dit
isd
enote
dby
0.
A=
00···
00
0···
0. . .
. . .. .
.. . .
00···
0 E
xam
ple
:
( 00
00
00) is
anu
llm
atr
ixof
ord
er2×
3.
4.
Up
per
tria
ngu
lar
matr
ix:
Th
esq
uare
matr
ixA
=(aij
)of
ord
ern
isca
lled
an
up
per
tria
ngu
lar
matr
ixifaij
=0
for
alli>j,
an
dit
is
wri
tten
as
A=
a11
a12
a13···
a1n
0a22
a23···
a2n
0a32
a33···
a3n
. . .. . .
. ..
. . .0
00···ann
E
xam
ple
:
23
10−
14
00
5 isan
up
per
tria
ngula
rm
atr
ixof
ord
er3.
![Page 11: (C) - KSUfac.ksu.edu.sa/sites/default/files/mdhkr_qdym_lmqrr_104...c) ertices: V (0 b), V 2 (0; b) 2 b. 2 a ts: W 1 (a; 0), W 2 (a; 0). 7 aph. (a) 9 x 2 25 y 2 225 (b) 16 x 2 9 y 2](https://reader035.fdocuments.us/reader035/viewer/2022062415/5fdb545c9fe12d2da100204e/html5/thumbnails/11.jpg)
M-1
04
GE
NE
RA
LM
AT
HE
MA
TIC
S2
CH
AP
TE
R2:
Matr
ices
an
dD
ete
rmin
ants
Date
:/
/D
ay:
Page:
09
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
5.L
ow
er
tria
ngu
lar
matr
ix:
Th
esq
uar
em
atr
ixA
=(aij
)of
ord
ern
isca
lled
alo
wer
tria
ngu
lar
mat
rix
ifaij
=0
for
alli<j
,an
dit
is
wri
tten
asA
=
a11
00···
0a21
a22
0···
0a31
a32
a33···
0. . .
. . .. .
.. . .
an1
an2
an3···ann
E
xam
ple
:
40
01−
10
23
5 isa
low
ertr
ian
gula
rm
atr
ixof
ord
er3.
6.D
iagon
al
matr
ix:
Th
esq
uar
em
atri
xA
=(aij
)of
ord
ern
isca
lled
ad
iago
nal
mat
rix
ifaij
=0
for
allij
,an
dit
isw
ritt
enas
A=
a11
00···
00
a22
0···
00
0a33···
0. . .
. . .. .
.. . .
00
0···ann
E
xam
ple
:
20
00
10
00
5
isa
dia
gon
alm
atri
xof
ord
er3.
7.Id
enti
tym
atr
ix:
Th
esq
uar
em
atri
xI n
=(aij
)of
ord
ern
isca
lled
anid
enti
tym
atri
xifaij
=0
for
alli
=j
andaij
=1
for
alli
=j,
an
dit
is
I n=
10
0···
00
10···
00
01···
0. . .
. . .. .
.. . .
00
0···
1 E
xam
ple
:
10
00
10
00
1
isan
iden
tity
mat
rix
of
ord
er3.
Matr
ixO
pera
tion
s:(1
)A
dd
itio
nan
dsu
btr
acti
on
of
matr
ices
:A
dd
itio
nor
sub
tract
ion
of
two
matr
ices
isd
efin
edif
the
two
matr
ices
hav
eth
esa
me
ord
er.
IfA
=(aij
) m×n
an
dB
=(bij
) m×n
any
two
matr
ices
of
ord
erm×n
then
1.A
+B
=(aij
+b ij) m×n.
A+
B=
a11
+b 1
1a12
+b 1
2···
a1n
+b 1n
a21
+b 2
1a22
+b 2
2···
a2n
+b 2n
. . .. . .
. ..
. . .am
1+b m
1am
2+b m
2···amn
+b m
n
2.A−B
=(aij−b ij) m×n.
A-
B=
a11−b 1
1a12−b 1
2···
a1n−b 1n
a21−b 2
1a22−b 2
2···
a2n−b 2n
. . .. . .
. ..
. . .am
1−b m
1am
2−b m
2···amn−b m
n
E
xam
ple
19
IfA
=
13
25−
46
09
2 and
B=
50
81
4−
110
11−
2
F
indA
+B
an
dA−B
.
Note
s:1.
Th
ead
dit
ion
of
matr
ices
isco
mm
uta
tive
:ifA
an
dB
any
two
matr
ices
of
the
sam
eord
erth
enA
+B
=B
+A
.2.
Th
enu
llm
atr
ixis
the
iden
tity
elem
ent
of
ad
dit
ion
:ifA
isany
matr
ixth
enA
+0
=A
.
![Page 12: (C) - KSUfac.ksu.edu.sa/sites/default/files/mdhkr_qdym_lmqrr_104...c) ertices: V (0 b), V 2 (0; b) 2 b. 2 a ts: W 1 (a; 0), W 2 (a; 0). 7 aph. (a) 9 x 2 25 y 2 225 (b) 16 x 2 9 y 2](https://reader035.fdocuments.us/reader035/viewer/2022062415/5fdb545c9fe12d2da100204e/html5/thumbnails/12.jpg)
M-1
04
GE
NE
RA
LM
AT
HE
MA
TIC
S2
CH
AP
TE
R2:
Matr
ices
an
dD
ete
rmin
ants
Date
:/
/D
ay:
Page:
10
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
(2)
Mu
ltip
lyin
ga
matr
ixby
asc
ala
r:IfA
=(aij
)is
am
atri
xof
ord
erm×n
andc∈R
then
cA
=(caij
).
cA
=
ca11
ca12···
ca1n
ca21
ca22···
ca2n
. . .. . .
. ..
. . .cam
1cam
2···camn
E
xam
ple
20
IfA
=
( 13
20
92) ,
fin
d3A
.
Exam
ple
21
IfA
=
( 16
−2
4) an
dB
=
( 23
08
) ,fi
nd−
2A+
3B
.
(3)
Mu
ltip
lyin
ga
row
vecto
rby
acolu
mn
vecto
r:
IfA
=(a
1a2...an)
isa
row
vect
orof
ord
ern
an
dB
=
b 1 b 2 . . . b n
is
aco
lum
nve
ctor
ofor
dern
then
AB
=(a
1a2...an)B
=
b 1 b 2 . . . b n
=
a1b 1
+a2b 2
+...+
anb n
Exam
ple
22
IfA
=(2
14)
an
dB
=
8 −3 5
,fin
dAB
Exam
ple
23
IfA
=(2
14−
6)an
dB
=
1 3 −4 7
,fin
dAB
(4)
Mu
ltip
licati
on
of
matr
ices:
1.
IfA
an
dB
any
two
matr
ices
then
AB
isd
efin
edif
the
nu
mb
erof
colu
mn
sofA
equ
als
the
nu
mb
erof
row
sofB
.
2.
IfA
=(aij
) m×n
an
dB
=(bij
) n×
then
AB
=(cij
) m×p.
c ij
isca
lcu
late
dby
mu
ltip
lyin
gth
eith
row
ofA
by
thejth
colu
mn
of
B.
Exam
ple
24
IfA
=
( 16
−2
4
) an
dB
=
( 23
08) ,
fin
dAB
.
Exam
ple
25
IfA
=
( 16
2−
24
1
) an
dB
=
23
0−
14
20
17 ,
fin
dAB
.
Exam
ple
26
IfA
=( 1
35) ,
B=
2 −1 0
and
C=
23
−1
40
1 ,fi
nd
1.AB
2.BC
.
![Page 13: (C) - KSUfac.ksu.edu.sa/sites/default/files/mdhkr_qdym_lmqrr_104...c) ertices: V (0 b), V 2 (0; b) 2 b. 2 a ts: W 1 (a; 0), W 2 (a; 0). 7 aph. (a) 9 x 2 25 y 2 225 (b) 16 x 2 9 y 2](https://reader035.fdocuments.us/reader035/viewer/2022062415/5fdb545c9fe12d2da100204e/html5/thumbnails/13.jpg)
M-1
04
GE
NE
RA
LM
AT
HE
MA
TIC
S2
CH
AP
TE
R2:
Matr
ices
an
dD
ete
rmin
ants
Date
:/
/D
ay:
Page:
11
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
Note
s:1
Matr
ixm
ult
ipli
cati
on
isnot
com
mu
tati
ve.
Exam
ple
27
IfA
=
( 43
9−
12
0) an
dB
=
( 23
01
) Com
pu
te(i
fpo
s-
sibl
e)1.AB
2.BA
Exam
ple
28
IfA
=
( 34
−1
2) an
dB
=
( 75
10) ,
fin
d1.AB
2.
BA
.
2T
he
identi
tym
atr
ixis
the
identi
tyele
ment
inm
atr
ixm
ul-
tip
licati
on
.
I n=
10···
00
1···
00
01···
0. . .
. . .. .
.. . .
00···
1
IfA
isa
mat
rix
ofor
derm×n
andI n
isth
eid
enti
tym
atr
ixof
ord
ern
then
AI n
=I nA
=A.
Exam
ple
29
IfA
=
( 43
−1
2) ,fi
ndAI 2
Pro
pert
ies
of
op
era
tion
son
matr
ices:
1.
IfA
,B
an
dC
any
thre
em
atr
ices
of
the
sam
eord
erth
en
A+B
+C
=(A
+B
)+C
=A
+(B
+C
)=
(A+C
)+B
2.
IfA
,B
any
two
matr
ices
of
ord
erm×n
an
dC
am
atr
ixof
ord
ern×p
then
(A+B
)C=AC
+BC
3.
IfA
,B
any
two
matr
ices
of
ord
erm×n
an
dC
am
atr
ixof
ord
erp×m
then
C(A
+B
)=CA
+CB
4.
IfA
am
atr
ixof
ord
erm×n
,B
am
atr
ixof
ord
ern×p
an
dC
am
atr
ixof
ord
erp×q
then
ABC
=(AB
)C=A
(BC
)
Tra
nsp
ose
of
am
atr
ix:
IfA
=(aij
) m×n
then
the
tran
spose
ofA
isAt
=(aji)n×m
.
Exam
ple
30
IfA
=
( 43
−1
2) ,fi
ndAt
Note
:T
he
tran
spose
of
alo
wer
tria
ngu
lar
matr
ixis
an
up
per
tria
ngu
-la
rm
atr
ix,
an
dth
etr
an
spose
of
an
up
per
tria
ngu
lar
matr
ixis
alo
wer
tria
ngu
lar
matr
ix.
Pro
pert
ies
of
tran
spose
of
am
atr
ix:
IfA
an
dB
any
two
matr
ices
an
dλ∈R
then
1.
(At)t
=A
.2.
(A+B
)t=At
+Bt
.3.
(λA
)t=λAt
.4.
(AB
)t=BtAt
.
![Page 14: (C) - KSUfac.ksu.edu.sa/sites/default/files/mdhkr_qdym_lmqrr_104...c) ertices: V (0 b), V 2 (0; b) 2 b. 2 a ts: W 1 (a; 0), W 2 (a; 0). 7 aph. (a) 9 x 2 25 y 2 225 (b) 16 x 2 9 y 2](https://reader035.fdocuments.us/reader035/viewer/2022062415/5fdb545c9fe12d2da100204e/html5/thumbnails/14.jpg)
M-1
04
GE
NE
RA
LM
AT
HE
MA
TIC
S2
CH
AP
TE
R2:
Matr
ices
an
dD
ete
rmin
ants
Date
:/
/D
ay:
Page:
12
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
(2)
Dete
rmin
ants
IfA
isa
squ
are
mat
rix
then
the
det
erm
inan
tofA
isd
enote
dbydet
(A)
or|A|.
(A)
Th
ed
ete
rmin
ant
of
a2×
2m
atr
ix:
IfA
=
( a 11a12
a21
a22
) ,th
endet
(A)
=
∣ ∣ ∣ ∣a 11a12
a21
a22
∣ ∣ ∣ ∣=a11a22−a21a12
Exam
ple
31
fin
ddet
(A):
(1)
A=
( 15
37) (2
)A
=
( 4−
12
9
)
(B)
Th
ed
ete
rmin
ant
of
a3×
3m
atr
ix:
IfA
=
a 11a12
a13
a21
a22
a23
a31
a32
a33
,th
en
det
(A)
=a11
∣ ∣ ∣ ∣a 22a23
a32
a33
∣ ∣ ∣ ∣−a12
∣ ∣ ∣ ∣a 21a23
a31
a33
∣ ∣ ∣ ∣+a13
∣ ∣ ∣ ∣a 21a22
a31
a32
∣ ∣ ∣ ∣=a11(a
22a33−a23a32)−a12(a
21a33−a23a31)
+a13(a
21a32−a22a31)
Exam
ple
32
Fin
d|A|:
(1)
A=
16
35−
14
−2
97 (2
)A
=
41
52
1−
21
87
(C)
Th
ed
ete
rmin
ant
of
a4×
4m
atr
ix:
IfA
=
a11
a12
a13
a14
a21
a22
a23
a24
a31
a32
a33
a34
a41
a42
a43
a44
,th
en
det
(A)
=a11det
(A1)−a12det
(A2)
+a13det
(A3)−a14det
(A4)
wh
ere
A1
=
a 22a23
a24
a32
a33
a34
a42
a43
a44
A 2=
a 21a23
a24
a31
a33
a34
a41
a43
a44
A
3=
a 21a22
a24
a31
a32
a34
a41
a42
a44
A 4=
a 21a22
a23
a31
a32
a33
a41
a42
a43
E
xam
ple
33
Fin
ddet
(A):
(1)
A=
16
32
5−
14
1−
29
73
71
3−
6
(2)
A=
41
52
21
01
01
39
17
46
![Page 15: (C) - KSUfac.ksu.edu.sa/sites/default/files/mdhkr_qdym_lmqrr_104...c) ertices: V (0 b), V 2 (0; b) 2 b. 2 a ts: W 1 (a; 0), W 2 (a; 0). 7 aph. (a) 9 x 2 25 y 2 225 (b) 16 x 2 9 y 2](https://reader035.fdocuments.us/reader035/viewer/2022062415/5fdb545c9fe12d2da100204e/html5/thumbnails/15.jpg)
M-1
04
GE
NE
RA
LM
AT
HE
MA
TIC
S2
CH
AP
TE
R2:
Matr
ices
an
dD
ete
rmin
ants
Date
:/
/D
ay:
Page:
13
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
Pro
pert
ies
of
dete
rmin
ants
:1.
IfA
isa
squ
are
mat
rix
that
conta
ins
aze
roro
w(o
ra
zero
colu
mn
)th
endet
(A)
=0.
2.IfA
isa
squ
are
mat
rix
that
conta
ins
two
equ
al
row
s(o
rtw
oeq
ual
colu
mn
s)th
endet
(A)
=0.
3.IfA
isa
squ
are
mat
rix
that
conta
ins
aro
ww
hic
his
am
ult
iple
of
anot
her
row
(or
aco
lum
nw
hic
his
am
ult
iple
of
an
oth
erco
lum
n)
then
det
(A)
=0.
4.IfA
isa
dia
gon
alm
atri
xor
anup
per
tria
ngu
lar
matr
ixor
alo
wer
tria
ngu
lar
mat
rix
thedet
(A)
isth
eth
ep
rod
uct
of
the
elem
ents
of
the
mai
nd
iago
nal
.
5.T
he
det
erm
inan
tof
the
nu
llm
atri
xis
0an
dth
ed
eter
min
ant
of
the
iden
tity
mat
rix
is1.
6.IfA
isa
squ
are
mat
rix
andB
isth
em
atri
xfo
rmed
by
mu
ltip
lyin
gon
eof
the
row
s(o
rco
lum
ns)
ofA
by
an
on
-zer
oco
nst
antλ
then
det
(B)
=λdet
(A)
.
7.IfA
isa
squ
are
mat
rix
andB
isth
em
atri
xfo
rmed
by
inte
rch
an
g-
ing
two
row
s(o
rtw
oco
lum
ns)
ofA
then
det
(B)
=−det
(A).
8.IfA
isa
squ
are
mat
rix
andB
isth
em
atri
xfo
rmed
by
mu
ltip
lyin
ga
row
by
an
on-z
ero
const
ant
and
add
ing
the
resu
ltto
an
oth
erro
w(o
rm
ult
iply
ing
aco
lum
nby
an
on-z
ero
const
ant
an
dad
din
gth
ere
sult
toan
oth
erco
lum
n)
then
det
(B)
=det
(A).
Exam
ple
34
Use
pro
pert
ies
of
det
erm
inan
tsto
calc
ula
teth
edet
erm
i-n
an
tsof
the
foll
ow
ing
matr
ices
(1)
A=
12−
20
00
34
7
(2)
A=
12
16
56
34
3 (3
)A
=
12−
24
75
36−
6 (4)
A=
30
46−
12
00
5 (5
)A
=
52
315
81
10
62
(6)
A=
12
34
02
30
12
35
30
00
![Page 16: (C) - KSUfac.ksu.edu.sa/sites/default/files/mdhkr_qdym_lmqrr_104...c) ertices: V (0 b), V 2 (0; b) 2 b. 2 a ts: W 1 (a; 0), W 2 (a; 0). 7 aph. (a) 9 x 2 25 y 2 225 (b) 16 x 2 9 y 2](https://reader035.fdocuments.us/reader035/viewer/2022062415/5fdb545c9fe12d2da100204e/html5/thumbnails/16.jpg)
M-1
04
GE
NE
RA
LM
AT
HE
MA
TIC
S2
CH
AP
TE
R2:
Matr
ices
an
dD
ete
rmin
ants
Date
:/
/D
ay:
Page:
14
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
Hom
ew
ork
(2):
Qu
est
ion
1:
IfA
=
13
25−
46
09
2 ,B=
50
14
1011
and
C=
−20
07
53
Fin
d 1.B
+C
2.2B
+3C
3.C−B
4.A−C
5.AB
6.BA
7.At
8.(3A
)t
9.det
(A)
10.det
(2A
)
Qu
est
ion
2:
IfB
=7A
an
ddet
(A)
=−
3,
fin
ddet
(B).
Qu
est
ion
3:
Fin
dth
efo
llow
ing
det
erm
inants
:
1.
∣ ∣ ∣ ∣1−
22
7
∣ ∣ ∣ ∣
2.
∣ ∣ ∣ ∣ ∣ ∣1−
23
40
12
70∣ ∣ ∣ ∣ ∣ ∣
![Page 17: (C) - KSUfac.ksu.edu.sa/sites/default/files/mdhkr_qdym_lmqrr_104...c) ertices: V (0 b), V 2 (0; b) 2 b. 2 a ts: W 1 (a; 0), W 2 (a; 0). 7 aph. (a) 9 x 2 25 y 2 225 (b) 16 x 2 9 y 2](https://reader035.fdocuments.us/reader035/viewer/2022062415/5fdb545c9fe12d2da100204e/html5/thumbnails/17.jpg)
M-1
04
GE
NE
RA
LM
AT
HE
MA
TIC
S2
CH
AP
TE
R3:
SY
ST
EM
SO
FL
INE
AR
EQ
UA
TIO
NS
Date
:/
/D
ay:
Page:
15
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
Con
sid
erth
esy
stem
ofli
nea
req
uat
ion
sin
2d
iffer
ent
vari
ab
les
a11x1
+a12x2
=b 1
a21x1
+a22x2
=b 2
Th
eab
ove
syst
emof
lin
ear
equ
atio
ns
can
be
wri
tten
as
:AX
=B
wh
ere
A=
( a 11a12
a21
a22
) ,X
=
( x 1 x2
) ,an
dB
=
( b 1 b 2
) .
Gen
eral
ly,co
nsi
der
the
syst
emof
lin
ear
equ
atio
ns
inn
diff
eren
tva
riab
les
a11x1
+a12x2
+a13x3
+...+
a1nxn
=b 1
a21x1
+a22x2
+a23x3
+...+
a2nxn
=b 2
a31x1
+a32x2
+a33x3
+...+
a3nxn
=b 3
. . .+. . .+
. . .+. . .+
. . .=
. . .
an1x1
+an2x2
+an3x3
+...+
+annxn
=b n
(1)
Th
eab
ove
syst
emof
lin
ear
equ
atio
ns
can
be
wri
tten
as
:AX
=B
wh
ere
A=
a11
a12···
a1n
a21
a22···
a2n
. . .. . .
. ..
. . .an1
an2···ann
,X
=
x1
x2 . . . xn
,an
dB
=
b 1 b 2 . . . b n
.
Ais
call
edth
eco
effici
ents
mat
rix
Xis
call
edth
eco
lum
nve
ctor
ofva
riab
les
(or
colu
mn
vec
tor
of
the
un
kn
own
s)B
isca
lled
the
colu
mn
vec
tor
ofco
nst
ants
(or
colu
mn
vec
tor
of
the
resu
ltan
ts)
Th
eore
m1
The
syst
emof
lin
ear
equ
ati
on
s(3
)has
aso
luti
on
ifdet
(A)6=
0.
Th
isch
ap
ter
pre
sents
thre
em
eth
od
sof
solv
ing
the
syst
emof
lin
ear
equ
ati
on
s(3
),th
efirs
tm
eth
od
isC
ram
ers
rule
,th
ese
con
dis
Gau
sseli
min
ati
on
meth
od
,an
dth
eth
ird
isG
au
ss-J
ord
an
meth
od
.
(1)
Cra
mers
Ru
leC
on
sid
erth
esy
stem
of
lin
ear
equ
ati
on
sinn
diff
eren
tva
riab
les
(3).
Th
em
eth
od
:Ifdet
(A)6=
0,
then
the
solu
tion
of
the
syst
em(3
)is
giv
enby
xi
=det
(Ai)
det
(A)
for
ever
yi
=1,
2,...,n.
wh
ereAi
isth
em
atr
ixfo
rmed
by
rep
laci
ng
theith
colu
mn
ofA
by
the
colu
mn
vec
tor
of
con
stants
:
A1
=
b 1a12···
a1n
b 2a22···
a2n
. . .. . .
. ..
. . .b n
an2···ann
,A2
=
a11
b 1···
a1n
a21
b 2···
a2n
. . .. . .
. ..
. . .an1
b n···ann
,...,An
=
a11
a12···
b 1a21
a22···
b 2. . .
. . .. .
.. . .
an1
an2···b n
,
Exam
ple
35
Use
Cra
mer
sru
leto
solv
eth
esy
stem
of
lin
ear
equ
ati
on
s
2x
+3y
=7
−x
+y
=4
![Page 18: (C) - KSUfac.ksu.edu.sa/sites/default/files/mdhkr_qdym_lmqrr_104...c) ertices: V (0 b), V 2 (0; b) 2 b. 2 a ts: W 1 (a; 0), W 2 (a; 0). 7 aph. (a) 9 x 2 25 y 2 225 (b) 16 x 2 9 y 2](https://reader035.fdocuments.us/reader035/viewer/2022062415/5fdb545c9fe12d2da100204e/html5/thumbnails/18.jpg)
M-1
04
GE
NE
RA
LM
AT
HE
MA
TIC
S2
CH
AP
TE
R3:
SY
ST
EM
SO
FL
INE
AR
EQ
UA
TIO
NS
Date
:/
/D
ay:
Page:
16
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
Exam
ple
36
Use
Cra
mer
sru
leto
solv
eth
esy
stem
of
lin
ear
equ
ati
on
s
2x+y
+z
=3
4x+y−z
=−
2
2x−
2y+z
=6
Exam
ple
37
Use
Cra
mer
sru
leto
solv
eth
esy
stem
of
lin
ear
equ
ati
on
s
x1
+2x2
=1
2x1
+x2
=−
1
Exam
ple
38
Use
Cra
mer
sru
leto
solv
eth
esy
stem
of
lin
ear
equ
ati
on
s
x+y
+z
=12
x−y
=2
x−z
=4
![Page 19: (C) - KSUfac.ksu.edu.sa/sites/default/files/mdhkr_qdym_lmqrr_104...c) ertices: V (0 b), V 2 (0; b) 2 b. 2 a ts: W 1 (a; 0), W 2 (a; 0). 7 aph. (a) 9 x 2 25 y 2 225 (b) 16 x 2 9 y 2](https://reader035.fdocuments.us/reader035/viewer/2022062415/5fdb545c9fe12d2da100204e/html5/thumbnails/19.jpg)
M-1
04
GE
NE
RA
LM
AT
HE
MA
TIC
S2
CH
AP
TE
R3:
SY
ST
EM
SO
FL
INE
AR
EQ
UA
TIO
NS
Date
:/
/D
ay:
Page:
17
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
(2)
Gau
sseli
min
ati
on
meth
od
Con
sid
erth
esy
stem
ofli
nea
req
uat
ion
sinn
diff
eren
tva
riab
les:
a11x1
+a12x2
+a13x3
+...+
a1nxn
=b 1
a21x1
+a22x2
+a23x3
+...+
a2nxn
=b 2
a31x1
+a32x2
+a33x3
+...+
a3nxn
=b 3
. . .+. . .+
. . .+. . .+
. . .=
. . .
an1x1
+an2x2
+an3x3
+...+
+annxn
=b n
(2)
AX
=B
wh
ere
A=
a11
a12···
a1n
a21
a22···
a2n
. . .. . .
. ..
. . .an1
an2···ann
,X
=
x1
x2 . . . xn
,an
dB
=
b 1 b 2 . . . b n
.
Th
em
eth
od
:
1.C
onst
ruct
the
augm
ente
dm
atri
x[A|B
].
a11
a12···
a1n
b 1a21
a22···
a2n
b 2. . .
. . .. .
.. . .
. . .an1
an2···ann
b n
2.
Use
elem
enta
ryro
wop
erat
ion
son
the
augm
ente
dm
atr
ixto
tran
sform
the
mat
rixA
toan
up
per
tria
ngu
lar
mat
rix
wit
hle
ad
ing
coeffi
cien
tof
each
row
equ
als
1. 1
c 12
c 13
c 14···
c 1n
d1
01
c 33
c 24···
c 2n
d2
. . .. . .
. ..
. . .. . .
00
0···
1c (n−1)n
dn−1
00
00···
1dn
3.
Fro
mth
ela
stau
gmen
ted
mat
rix,xn
=dn
and
the
rest
ofth
eu
nkn
own
sca
nb
eca
lcu
late
dby
back
ward
sub
stit
u-
tion
.
Exam
ple
39
Use
Gau
ssel
imin
ati
on
met
hod
toso
lve
the
syst
em
x−
2y+z
=4
−x
+2y
+z
=−
2
4x−
3y−z
=−
4
![Page 20: (C) - KSUfac.ksu.edu.sa/sites/default/files/mdhkr_qdym_lmqrr_104...c) ertices: V (0 b), V 2 (0; b) 2 b. 2 a ts: W 1 (a; 0), W 2 (a; 0). 7 aph. (a) 9 x 2 25 y 2 225 (b) 16 x 2 9 y 2](https://reader035.fdocuments.us/reader035/viewer/2022062415/5fdb545c9fe12d2da100204e/html5/thumbnails/20.jpg)
M-1
04
GE
NE
RA
LM
AT
HE
MA
TIC
S2
CH
AP
TE
R3:
SY
ST
EM
SO
FL
INE
AR
EQ
UA
TIO
NS
Date
:/
/D
ay:
Page:
18
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
Exam
ple
40
Use
Gau
ssel
imin
ati
on
met
hod
toso
lve
the
syst
em
x+y
+z
=2
x−y
+2z
=0
2x+z
=2
Exam
ple
41
Use
Gau
ssel
imin
ati
on
met
hod
toso
lve
the
syst
em
3x1
+x2
=9
x1
+2x2
=8
Exam
ple
42
Use
Gau
ssel
imin
ati
on
met
hod
toso
lve
the
syst
em
x+
2y
+3z
=14
2x
+y
+2z
=10
3x+
4y−
3z=
2
![Page 21: (C) - KSUfac.ksu.edu.sa/sites/default/files/mdhkr_qdym_lmqrr_104...c) ertices: V (0 b), V 2 (0; b) 2 b. 2 a ts: W 1 (a; 0), W 2 (a; 0). 7 aph. (a) 9 x 2 25 y 2 225 (b) 16 x 2 9 y 2](https://reader035.fdocuments.us/reader035/viewer/2022062415/5fdb545c9fe12d2da100204e/html5/thumbnails/21.jpg)
M-1
04
GE
NE
RA
LM
AT
HE
MA
TIC
S2
CH
AP
TE
R3:
SY
ST
EM
SO
FL
INE
AR
EQ
UA
TIO
NS
Date
:/
/D
ay:
Page:
19
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
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==
==
==
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==
==
==
(3)
Gau
ss-J
ord
an
meth
od
Con
sid
erth
esy
stem
ofli
nea
req
uat
ion
sinn
diff
eren
tva
riab
les:
a11x1
+a12x2
+a13x3
+...+
a1nxn
=b 1
a21x1
+a22x2
+a23x3
+...+
a2nxn
=b 2
a31x1
+a32x2
+a33x3
+...+
a3nxn
=b 3
. . .+. . .+
. . .+. . .+
. . .=
. . .
an1x1
+an2x2
+an3x3
+...+
+annxn
=b n
(3)
AX
=B
wh
ere
A=
a11
a12···
a1n
a21
a22···
a2n
. . .. . .
. ..
. . .an1
an2···ann
,X
=
x1
x2 . . . xn
,an
dB
=
b 1 b 2 . . . b n
.
Th
em
eth
od
:
1.C
onst
ruct
the
augm
ente
dm
atri
x[A|B
].
a11
a12···
a1n
b 1a21
a22···
a2n
b 2. . .
. . .. .
.. . .
. . .an1
an2···ann
b n
2.
Use
elem
enta
ryro
wop
erat
ion
son
the
augm
ente
dm
atr
ixto
tran
sform
the
mat
rixA
toth
eid
enti
tym
atri
x.
10
00···
0d1
01
00···
0d2
. . .. . .
. ..
. . .. . .
00
0···
10
dn−1
00
00···
1dn
3.
Fro
mth
ela
stau
gmen
ted
mat
rix,xi
=di
for
ever
yi
=1,
2,,n
.
Exam
ple
43
Use
Gau
ss-J
ord
an
met
hod
toso
lve
the
syst
em
x+y
=2
2x+y
=1
![Page 22: (C) - KSUfac.ksu.edu.sa/sites/default/files/mdhkr_qdym_lmqrr_104...c) ertices: V (0 b), V 2 (0; b) 2 b. 2 a ts: W 1 (a; 0), W 2 (a; 0). 7 aph. (a) 9 x 2 25 y 2 225 (b) 16 x 2 9 y 2](https://reader035.fdocuments.us/reader035/viewer/2022062415/5fdb545c9fe12d2da100204e/html5/thumbnails/22.jpg)
M-1
04
GE
NE
RA
LM
AT
HE
MA
TIC
S2
CH
AP
TE
R3:
SY
ST
EM
SO
FL
INE
AR
EQ
UA
TIO
NS
Date
:/
/D
ay:
Page:
20
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==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
Exam
ple
44
Use
Gau
ss-J
ord
an
met
hod
toso
lve
the
syst
em
x+y
+z
=2
x−y
+2z
=0
2x+z
=2
Exam
ple
45
Use
Gau
ss-J
ord
an
met
hod
toso
lve
the
syst
em
x−
2y+
2z
=5
5x+
3y
+6z
=57
x+
2y
+2z
=21
![Page 23: (C) - KSUfac.ksu.edu.sa/sites/default/files/mdhkr_qdym_lmqrr_104...c) ertices: V (0 b), V 2 (0; b) 2 b. 2 a ts: W 1 (a; 0), W 2 (a; 0). 7 aph. (a) 9 x 2 25 y 2 225 (b) 16 x 2 9 y 2](https://reader035.fdocuments.us/reader035/viewer/2022062415/5fdb545c9fe12d2da100204e/html5/thumbnails/23.jpg)
M-1
04
GE
NE
RA
LM
AT
HE
MA
TIC
S2
CH
AP
TE
R3:
SY
ST
EM
SO
FL
INE
AR
EQ
UA
TIO
NS
Date
:/
/D
ay:
Page:
21
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
Hom
ew
ork
(3):
Qu
est
ion
1:
Use
Cra
mer
sru
leto
solv
eth
esy
stem
of
lin
ear
equ
ati
on
s1.
x+y
+z
=18
x−y
+z
=6
x+y−z
=4
2.
2x−
4y
+3z
=10
3x+y−
2z=
6
x+
3y−z
=20
Qu
est
ion
2:
Use
Gau
ssel
imin
atio
nm
eth
od
toso
lve
the
syst
em1.
x+y
+z
=18
x−y
+z
=6
x+y−z
=4
2.
x+y
+z
=12
x−y
=2
x−z
=4
Qu
est
ion
3:
Use
Gau
ss-J
ord
an
met
hod
toso
lve
the
syst
em1.
x−
3y+z
=21
4x+
2y
+z
=14
3x+
3y
+z
=7
2.
2x−
4y+
3z
=10
3x+y−
2z=
6
x+
3y−z
=20
![Page 24: (C) - KSUfac.ksu.edu.sa/sites/default/files/mdhkr_qdym_lmqrr_104...c) ertices: V (0 b), V 2 (0; b) 2 b. 2 a ts: W 1 (a; 0), W 2 (a; 0). 7 aph. (a) 9 x 2 25 y 2 225 (b) 16 x 2 9 y 2](https://reader035.fdocuments.us/reader035/viewer/2022062415/5fdb545c9fe12d2da100204e/html5/thumbnails/24.jpg)
M-1
04
GE
NE
RA
LM
AT
HE
MA
TIC
S2
CH
AP
TE
R4:
INT
EG
RA
TIO
ND
ate
:/
/D
ay:
Page:
22
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==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
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==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
Anti
-deri
vati
ves
&In
defi
nit
ein
tegra
l(1
)A
nti
-deri
vati
ves
Anti
-deri
vati
ves
Defi
nit
ion
5A
fun
ctio
nF
isca
lled
an
an
ti-d
eriv
ati
veoff
on
an
inte
rvalI
if
F′ (x
)=f
(x)
for
ever
yx∈I.
Exam
ple
46
1.
LetF
(x)
=x2
+3x
+1
an
df
(x)
=2x
+3.
Sin
ceF
′ (x)
=f
(x),
the
fun
ctio
nF
(x)
isan
an
ti-d
eriv
ati
veof
f(x
).
2.
LetG
(x)
=si
n(x
)+x
an
dg(x
)=
cos(x
)+
1.
We
know
thatG
′ (x)
=co
s(x
)+
1an
dth
ism
ean
sth
efu
nct
ion
G(x
)is
an
an
ti-d
eriv
ati
veofg(x
).
Th
eore
m2
Ifth
efu
nct
ion
sF
(x)
an
dG
(x)
are
an
ti-d
eriv
ati
ves
of
afu
nct
ionf
(x)
on
the
inte
rvalI
,th
ere
exis
tsa
con
stan
tc
such
thatG
(x)−F
(x)
=c.
Exam
ple
47
Letf
(x)
=2x
.T
he
fun
ctio
ns
F(x
)=x2
+2,
G(x
)=x2−
1 2,
H(x
)=x2−
3√2,
an
dm
an
yoth
erfu
nct
ion
sare
an
ti-d
eriv
ati
ves
of
afu
nct
ionf
(x).
Gen
-er
all
y,fo
rth
efu
nct
ionf
(x)
=2x
,th
efu
nct
ionF
(x)
=x2
+c
isth
ean
ti-d
eriv
ati
vew
her
ec
isan
arb
itra
ryco
nst
an
t.
Exam
ple
48
Fin
dth
ege
ner
al
form
of
the
an
ti-d
eriv
ati
veoff
(x)
=6x
5.
(2)
Ind
efi
nit
eIn
tegra
ls
Ind
efi
nit
eIn
tegra
ls
Defi
nit
ion
6L
etf
bea
con
tin
uou
sfu
nct
ion
on
an
inte
rvalI
.T
he
indefi
nit
ein
tegr
al
off
(x)
isth
ege
ner
al
an
ti-d
eriv
ati
veof
f(x
)onI
an
dsy
mbo
lize
dby
∫ f(x
)dx
=F
(x)
+c.
The
fun
ctio
nf
(x)
isca
lled
the
inte
gran
d,
the
sym
bol
∫ isth
e
inte
gral
sign
,x
isca
lled
the
vari
abl
eof
inte
grati
on
an
dc
isth
eco
nst
an
tof
inte
grati
on
.
(3)
Pro
pert
ies
of
Ind
efi
nit
eIn
tegra
lsL
etf
an
dg
be
inte
gra
ble
funct
ion
s,th
en
1.
∫ ( f(x
)±g(x
)) dx=
∫ f(x
)±∫ g
(x)dx.
2.
∫ kf
(x)dx
=k
∫ f(x
)dx
,w
her
ek
isa
con
stant
Basi
cIn
tegra
tion
Ru
les:
1)
∫ xndx
=xn+
1
n+1
+c
wh
eren6=−
1.
Sp
ecia
lca
se:
∫ 1dx
=x
+c.
Exam
ple
49
Eva
luate
the
foll
ow
ing
inte
grals
:
1.
∫ x+
3dx
2.
∫ 4x3
+2x
+1dx
![Page 25: (C) - KSUfac.ksu.edu.sa/sites/default/files/mdhkr_qdym_lmqrr_104...c) ertices: V (0 b), V 2 (0; b) 2 b. 2 a ts: W 1 (a; 0), W 2 (a; 0). 7 aph. (a) 9 x 2 25 y 2 225 (b) 16 x 2 9 y 2](https://reader035.fdocuments.us/reader035/viewer/2022062415/5fdb545c9fe12d2da100204e/html5/thumbnails/25.jpg)
M-1
04
GE
NE
RA
LM
AT
HE
MA
TIC
S2
CH
AP
TE
R4:
INT
EG
RA
TIO
ND
ate
:/
/D
ay:
Page:
23
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==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
2)T
rigo
nom
etri
cF
un
ctio
ns:
∫ cosdx
=si
nx
+c
∫ sinxdx
=−
cosx
+c
∫ sec2
xdx
=ta
nx
+c
∫ csc2
xdx
=−
cotx
+c
∫ secx
tanxdx
=se
cx
+c
∫ cscx
cotxdx
=−
cscx
+c
Exam
ple
50
Eva
luate
the
foll
ow
ing
inte
grals
:
1.
∫ 2si
nx
+3
cosxdx
2.
∫ √ x+
sec2xdx
3)N
atu
ral
Log
arit
hm
ican
dE
xp
onen
tial
Fu
nct
ion
s∫ 1 x
dx
=lnx
+c
∫ exdx
=ex
+c
Exam
ple
51
Eva
luate
the
foll
ow
ing
inte
grals
:
1.
∫ 2 x−
csc2xdx
2.
∫ 3ex
+1
1+x2dx
3.
∫ 1 x+
1 x2dx
4)
Inve
rse
Tri
gon
om
etri
cF
un
ctio
ns
∫1
√1−x2dx
=si
n−1x
+c
wh
ere|x|<
1∫
1
1+x2dx
=ta
n−1x
+c
∫1
x√x2−
1dx
=sec−
1x
+c
wh
ere|x|>
1
Exam
ple
52
Eva
luate
the
foll
ow
ing
inte
grals
:
1.
∫ 31+x2dx
2.
∫1
√1−x2
+1 √xdx
![Page 26: (C) - KSUfac.ksu.edu.sa/sites/default/files/mdhkr_qdym_lmqrr_104...c) ertices: V (0 b), V 2 (0; b) 2 b. 2 a ts: W 1 (a; 0), W 2 (a; 0). 7 aph. (a) 9 x 2 25 y 2 225 (b) 16 x 2 9 y 2](https://reader035.fdocuments.us/reader035/viewer/2022062415/5fdb545c9fe12d2da100204e/html5/thumbnails/26.jpg)
M-1
04
GE
NE
RA
LM
AT
HE
MA
TIC
S2
CH
AP
TE
R4:
INT
EG
RA
TIO
ND
ate
:/
/D
ay:
Page:
24
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
(4)
Defi
nit
eIn
tegra
ls:
Defi
nit
eIn
tegra
ls
Defi
nit
ion
7L
etf
bea
fun
ctio
ndefi
ned
on
acl
ose
din
terv
al
[a,b
].Iff
isin
tegr
abl
eon
that
inte
rval
an
dF
(x)
isth
ege
ner
al
an
ti-d
eriv
ati
ve,
the
defi
nit
ein
tegr
al
off
is
∫ b a
f(x
)dx
=[ F
(x)] b a
=F
(b)−F
(a).
The
nu
mbe
rsa
an
db
are
call
edth
eli
mit
sof
the
inte
grati
on
.
Exam
ple
53
Eva
luate
the
foll
ow
ing
inte
grals
:
1.
∫ 2 −1
2x
+1dx
2.
∫ 3 0
x2
+1dx
3.
∫ 2 1
1 √x3dx
4.
∫π 2
0
sin
(x)
+1dx
5.
∫ π π 4
sec2
(x)−
4dx
6.
∫ 1 0
2x
+exdx
7.
∫ 1 2
1 x+√xdx
![Page 27: (C) - KSUfac.ksu.edu.sa/sites/default/files/mdhkr_qdym_lmqrr_104...c) ertices: V (0 b), V 2 (0; b) 2 b. 2 a ts: W 1 (a; 0), W 2 (a; 0). 7 aph. (a) 9 x 2 25 y 2 225 (b) 16 x 2 9 y 2](https://reader035.fdocuments.us/reader035/viewer/2022062415/5fdb545c9fe12d2da100204e/html5/thumbnails/27.jpg)
M-1
04
GE
NE
RA
LM
AT
HE
MA
TIC
S2
CH
AP
TE
R4:
INT
EG
RA
TIO
ND
ate
:/
/D
ay:
Page:
25
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
(5)
Inte
gra
tion
by
Su
bst
itu
tion
:
Su
bst
itu
tion
Met
hod
Th
eore
m3
Letg
bea
diff
eren
tiabl
efu
nct
ion
on
the
inte
rvalI
wher
eth
eder
ivati
veis
con
tin
uou
s.L
etf
bea
con
tin
uou
son
an
inte
rvalI
invo
lves
the
ran
geof
the
fun
ctio
ng
.IfF
isan
an
ti-d
eriv
ati
veof
the
fun
ctio
nf
onI
,th
en
∫ f(g
(x))g
′ (x)dx
=F
(g(x
))+c,
x∈I.
Ste
ps
of
Inte
gra
tion
by
Su
bst
itu
tion
:F
orsi
mp
lici
ty,
the
sub
stit
uti
onm
eth
od
can
be
sum
mari
zed
inth
efo
l-lo
win
gst
eps:
Ste
p1:
Ch
oos
ea
new
vari
ableu
.S
tep
2:
Det
erm
ine
the
valu
eofdu
.S
tep
3:
Mak
eth
esu
bst
itu
tion
i.e.
,el
imin
ate
all
occ
urr
ence
sofx
inth
ein
tegr
alby
mak
ing
the
enti
rein
tegr
alis
inte
rms
ofu
.S
tep
4:
Eva
luat
eth
en
ewin
tegr
al.
Ste
p5:
Ret
urn
the
eval
uat
ion
toth
ein
itia
lva
riablex
.
Exam
ple
54
Eva
luate
the
inte
gral
∫ (x+
1)3dx
.
Exam
ple
55
Eva
luate
the
inte
gral
∫ 2x(x
2+
1)3dx
.
Exam
ple
56
Eva
luate
the
inte
gral
∫ x2√ x3
+1dx
.
Exam
ple
57
Eva
luate
the
foll
ow
ing
inte
grals
:
1.
∫ 3co
s(3x
+4)dx
2.
∫ xse
c2(x
2)dx
3.
∫ cosxes
inxdx
4.
∫2x
x2
+1dx
![Page 28: (C) - KSUfac.ksu.edu.sa/sites/default/files/mdhkr_qdym_lmqrr_104...c) ertices: V (0 b), V 2 (0; b) 2 b. 2 a ts: W 1 (a; 0), W 2 (a; 0). 7 aph. (a) 9 x 2 25 y 2 225 (b) 16 x 2 9 y 2](https://reader035.fdocuments.us/reader035/viewer/2022062415/5fdb545c9fe12d2da100204e/html5/thumbnails/28.jpg)
M-1
04
GE
NE
RA
LM
AT
HE
MA
TIC
S2
CH
AP
TE
R4:
INT
EG
RA
TIO
ND
ate
:/
/D
ay:
Page:
26
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
(6)
Inte
gra
tion
by
Part
s:In
tegr
atio
nby
par
tsis
am
eth
od
totr
ansf
erth
eori
gin
al
inte
gra
lto
an
easi
erin
tegr
alth
atca
nb
eev
alu
ated
.P
ract
icall
y,th
ein
tegra
tion
by
par
tsd
ivid
esth
eor
igin
alin
tegr
alin
totw
op
art
su
an
ddv.
Th
en,
we
try
tofi
nd
thedu
by
der
ivin
gu
andv
by
inte
gra
tin
gdv.
Inte
gra
tion
by
Part
s
Th
eore
m4
Ifu
=f
(x)
an
dv
=g(x
)su
chth
atf′ (x
)an
dg′ (x
)are
con
tin
uou
s,th
en
∫ udv
=uv−∫ v
du.
Pro
of.
We
kn
owth
atd dx
(f(x
)g(x
))=f
(x)g′ (x
)+f′ (x
)g(x
).T
hu
s,
f(x
)g′ (x
)=
d dx
(f(x
)g(x
))−f′ (x
)g(x
).B
yin
tegr
atin
gth
eb
oth
sid
es,
we
hav
e
∫ f(x
)g′ (x
)dx
=
∫ d dx
(f(x
)g(x
))dx−∫ f
′ (x
)g(x
)dx
=f
(x)g
(x)−∫ f
′ (x
)g(x
)dx.
Sin
ceu
=f′ (x
)an
ddv
=g′ (x
),th
en
∫ udv
=uv−∫ v
du.�
Th
eore
m4
show
sth
atth
ein
tegr
atio
nby
par
tstr
an
sfer
sth
ein
tegra
l∫ u
dv
into
the
inte
gral
∫ vdu
that
shou
ldb
eea
sier
than
the
ori
gin
al
inte
gral
.T
he
ques
tion
her
eis
,w
hat
we
choose
asu
(x)
an
dw
hat
we
choos
eas
dv
=v′ (x
)dx
.It
isu
sefu
lto
chooseu
as
afu
nct
ion
that
sim
pli
fies
wh
end
iffer
enti
ated
,an
dto
choos
ev′
as
afu
nct
ion
that
sim
-p
lifi
esw
hen
inte
grat
ed.
Th
isis
exp
lain
edin
acl
eare
rsi
ght
thro
ugh
the
foll
owin
gex
amp
les.
Exam
ple
58
Eva
luate
the
foll
ow
ing
inte
gral
∫ xco
sxdx
.
Exam
ple
59
Eva
luate
the
foll
ow
ing
inte
gral
∫ xex
dx
.
Rem
ark
11.
Rem
embe
rth
at
when
we
con
sider
the
inte
grati
on
bypa
rts,
we
wan
tto
have
an
easi
erin
tegr
al.
As
we
saw
inE
xam
ple
59,
ifw
ech
ooseu
=ex
an
ddv
=xdx
we
have∫ x2 2
exdx
whic
his
more
diffi
cult
than
the
ori
gin
al
on
e.
2.
When
con
sider
ing
the
inte
grati
on
bypa
rts,
we
have
toch
oosedv
afu
nct
ion
that
can
bein
tegr
ate
d(s
eeE
xam
ple
60).
3.
Som
etim
esw
en
eed
tou
seth
ein
tegr
ati
on
bypa
rts
two
tim
esas
inE
xam
ple
s61
an
d62.
![Page 29: (C) - KSUfac.ksu.edu.sa/sites/default/files/mdhkr_qdym_lmqrr_104...c) ertices: V (0 b), V 2 (0; b) 2 b. 2 a ts: W 1 (a; 0), W 2 (a; 0). 7 aph. (a) 9 x 2 25 y 2 225 (b) 16 x 2 9 y 2](https://reader035.fdocuments.us/reader035/viewer/2022062415/5fdb545c9fe12d2da100204e/html5/thumbnails/29.jpg)
M-1
04
GE
NE
RA
LM
AT
HE
MA
TIC
S2
CH
AP
TE
R4:
INT
EG
RA
TIO
ND
ate
:/
/D
ay:
Page:
27
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
Exam
ple
60
Eva
luate
the
foll
ow
ing
inte
gral
∫ lnxdx
.
Exam
ple
61
Eva
luate
the
foll
ow
ing
inte
gral
∫ exco
sxdx
.
Exam
ple
62
Eva
luate
the
foll
ow
ing
inte
gral
∫ x2ex
dx
.
Exam
ple
63
Eva
luate
the
foll
ow
ing
inte
gral
∫ 1 0
tan−1xdx
.
![Page 30: (C) - KSUfac.ksu.edu.sa/sites/default/files/mdhkr_qdym_lmqrr_104...c) ertices: V (0 b), V 2 (0; b) 2 b. 2 a ts: W 1 (a; 0), W 2 (a; 0). 7 aph. (a) 9 x 2 25 y 2 225 (b) 16 x 2 9 y 2](https://reader035.fdocuments.us/reader035/viewer/2022062415/5fdb545c9fe12d2da100204e/html5/thumbnails/30.jpg)
M-1
04
GE
NE
RA
LM
AT
HE
MA
TIC
S2
CH
AP
TE
R4:
INT
EG
RA
TIO
ND
ate
:/
/D
ay:
Page:
28
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
(7)
Inte
gra
lsof
Rati
on
al
Fu
ncti
on
s:
We
stu
dy
rati
onal
fun
ctio
ns
ofth
efo
rmq(x
)=
f(x
)g(x
)w
her
ef
(x)
andg(x
)
are
pol
yn
omia
ls.
Th
ep
revio
us
tech
niq
ues
like
inte
gra
tion
by
part
sis
not
enou
ghto
eval
uat
eth
ein
tegr
alof
the
rati
onal
fun
ctio
ns.
Th
eref
ore
,w
en
eed
tokn
owa
new
tech
niq
ue
toh
elp
us
toev
alu
ate
the
inte
gra
lof
the
rati
onal
fun
ctio
ns.
Th
iste
chn
iqu
eis
call
edd
ecom
posi
tion
of
the
rati
onal
fun
ctio
ns
into
asu
mof
par
tial
frac
tion
s.
Th
ep
ract
ical
step
sof
inte
gral
sof
rati
onal
fun
ctio
ns
can
be
sum
mari
zed
asfo
llow
s:
Ste
p1:
Ifth
ed
egre
eofg(x
)is
less
than
the
deg
ree
off
(x),
we
do
pol
yn
omia
llo
ng-
div
isio
n,
oth
erw
ise
we
mov
eto
step
2.
Fro
mth
elo
ng
div
isio
nsh
own
onth
eri
ght
sid
e,w
eh
ave
q(x
)=f
(x)
g(x
)=h
(x)
+r(x
)
g(x
),
wh
ereh
(x)
isca
lled
the
qu
o-ti
ent
andr(x
)is
call
edth
ere
mai
nd
er.
h(x
)g(x
)) f(
x)
......
r(x)
Ste
p2:
Fac
tor
the
den
omin
atorg(x
)in
toir
red
uci
ble
poly
nom
ials
wh
ere
the
resu
ltis
eith
erli
nea
ror
irre
du
cib
lequ
adra
tic
poly
nom
ials
.
Ste
p3:
Fin
dth
ep
arti
alfr
acti
ond
ecom
pos
itio
n.
Th
isst
epd
epen
ds
on
step
2w
her
eif
deg
ree
off
(x)
isle
ssth
anth
ed
egre
eofg(x
),th
enth
e
frac
tion
f(x
)g(x
)ca
nb
ew
ritt
enas
asu
mof
par
tial
fract
ion
s:
q(x
)=P1(x
)+P2(x
)+P3(x
)+...+
Pn(x
),
wh
ere
eachPi(x
)=
A(ax+b)m,m∈N
orPi(x
)=
Ax+B
(ax2+bx+c)m
ifb2−
4ac<
0.T
he
con
stan
tsA,B,...
are
com
pu
ted
late
r.
Ste
p4:
Inte
grat
eth
ere
sult
ofst
ep3.
Exam
ple
64
Eva
luate
the
foll
ow
ing
inte
gral
∫x
+1
x2−
2x−
8dx
.
Exam
ple
65
Eva
luate
the
inte
gral
∫x
+3
(x−
3)(x−
2)dx
.
![Page 31: (C) - KSUfac.ksu.edu.sa/sites/default/files/mdhkr_qdym_lmqrr_104...c) ertices: V (0 b), V 2 (0; b) 2 b. 2 a ts: W 1 (a; 0), W 2 (a; 0). 7 aph. (a) 9 x 2 25 y 2 225 (b) 16 x 2 9 y 2](https://reader035.fdocuments.us/reader035/viewer/2022062415/5fdb545c9fe12d2da100204e/html5/thumbnails/31.jpg)
M-1
04
GE
NE
RA
LM
AT
HE
MA
TIC
S2
CH
AP
TE
R4:
INT
EG
RA
TIO
ND
ate
:/
/D
ay:
Page:
29
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
Exam
ple
66
Eva
luate
the
inte
gral
∫ 2x3−
4x2−
15x
+5
x2
+3x
+2
dx
.
Rem
ark
21.
The
nu
mbe
rof
con
stan
tsA,B,C,...
iseq
ual
toth
edeg
ree
of
the
den
om
inato
rg(x
).T
her
efore
,in
the
case
of
repe
ate
dfa
ctors
of
the
den
om
inato
r,w
ehave
toch
eck
the
nu
mbe
rof
the
con
stan
tsan
dth
edeg
ree
ofg(x
).
2.
Ifth
eden
om
inato
rg(x
)co
nta
ins
on
irre
du
cibl
equ
adra
tic
fact
ors
,th
en
um
erato
rsof
fract
ion
dec
om
posi
tion
shou
ldbe
poly
nom
ials
of
deg
ree
on
e.
Exam
ple
67
Eva
luate
the
foll
ow
ing
inte
gral
∫ 2x2−
25x−
33
(x+
1)2
(x−
5)dx
.
Exam
ple
68
Eva
luate
the
foll
ow
ing
inte
gral
∫ 2x2
+3x
+2
x3
+x
dx
.
![Page 32: (C) - KSUfac.ksu.edu.sa/sites/default/files/mdhkr_qdym_lmqrr_104...c) ertices: V (0 b), V 2 (0; b) 2 b. 2 a ts: W 1 (a; 0), W 2 (a; 0). 7 aph. (a) 9 x 2 25 y 2 225 (b) 16 x 2 9 y 2](https://reader035.fdocuments.us/reader035/viewer/2022062415/5fdb545c9fe12d2da100204e/html5/thumbnails/32.jpg)
M-1
04
GE
NE
RA
LM
AT
HE
MA
TIC
S2
CH
AP
TE
R4:
INT
EG
RA
TIO
ND
ate
:/
/D
ay:
Page:
30
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
Hom
ew
ork
(4):
Qu
est
ion
1:
Eva
luat
eth
efo
llow
ing
inte
gral
s:
1.
∫ sec2
(3x−
5)dx
2.
∫dx
√16−x2
3.
∫ xexdx
4.
∫ xco
sxdx
5.
∫ sin−1xdx
6.
∫dx
x2−x−
2
7.
∫ x(2x2−
3)8dx
8.
∫ cos
3√x
3√x2
dx
Qu
est
ion
2:
Eva
luate
the
foll
owin
gin
tegra
ls:
1.
∫ secx
+ta
nx
cosx
dx
2.
∫ xln√xdx
3.
∫ 3 1
x2
+1dx
4.
∫ 5 e
1
x−
2dx
5.
∫ 6 3
1
x−
2+
2
x+
1dx
6.
∫ π/2
0
(1+√
cosx
)2si
nxdx
![Page 33: (C) - KSUfac.ksu.edu.sa/sites/default/files/mdhkr_qdym_lmqrr_104...c) ertices: V (0 b), V 2 (0; b) 2 b. 2 a ts: W 1 (a; 0), W 2 (a; 0). 7 aph. (a) 9 x 2 25 y 2 225 (b) 16 x 2 9 y 2](https://reader035.fdocuments.us/reader035/viewer/2022062415/5fdb545c9fe12d2da100204e/html5/thumbnails/33.jpg)
M-1
04
GE
NE
RA
LM
AT
HE
MA
TIC
S2
CH
AP
TE
R5:
AP
PL
ICA
TIO
NS
OF
INT
EG
RA
TIO
ND
ate
:/
/D
ay:
Page:
31
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
(1)
Are
aIn
tegr
atio
nca
nb
eu
sed
toca
lcu
late
area
su
nd
erb
ou
nd
edgra
ph
s.In
gen
eral
,w
eco
nsi
der
the
follow
ing
case
s:1.
the
area
bet
wee
na
curv
e,th
ex-a
xis
or
y-a
xis
,an
dtw
ogiv
enor
din
ates
,2.
the
area
bet
wee
na
curv
e,th
ex-a
xis
ory-a
xis
,and
two
ord
inate
sgi
ven
by
cros
sin
gth
ecu
rve
the
axis
,3.
the
area
bet
wee
ntw
ocu
rves
.
1.Ify
=f
(x)
isco
n-
tinu
ous
on[a,b
]an
df
(x)≥
0∀x∈
[a,b
],th
ear
eaof
the
re-
gion
un
der
the
grap
hoff
(x)
from
x=a
tox
=b
isgi
ven
by
the
inte
gral
:
A=
∫ b a
f(x
)dx
xa
b
R
y
y=f
(x)
Iff
(x)
and
g(x
)ar
eco
nti
nu
ous
and
f(x
)≥g(x
)fo
rev
ery
x∈
[a,b
],th
enth
ear
eaA
ofth
ere
gion
bou
nd
edby
the
grap
hs
off
andg
isgi
ven
by
the
inte
gral
:
A=
∫ b a
( f(x)−g(x
)) dxa
b
f
g
R
x
y
3.
Ifx
=f
(y)
isco
n-
tinu
ou
son
[c,d
]an
df
(y)≥
0∀y∈
[c,d
],th
eare
aof
the
re-
gio
nu
nd
erth
egra
ph
off
(y)
fromy
=c
toy
=d
isgiv
enby
the
inte
gra
l:
A=
∫ d c
f(y
)dy
x
R
y abx
=f
(y)
4.
Iff
(y)
an
dg(y
)are
conti
nuou
san
df
(y)≥
g(y
)fo
rev
-er
yx∈
[c,d
],th
enth
eare
aA
of
the
re-
gio
nb
ou
nd
edby
the
gra
ph
soff
an
dg
isgiv
enby
the
inte
gra
l:
A=
∫ d c
( f(y)−g(y
)) dycd
f
g
R
x
y
Exam
ple
69
Ske
tch
the
regi
on
byth
egr
aph
ofy
=x
on
the
inte
rval
[0,3
],th
enfi
nd
its
are
a.
![Page 34: (C) - KSUfac.ksu.edu.sa/sites/default/files/mdhkr_qdym_lmqrr_104...c) ertices: V (0 b), V 2 (0; b) 2 b. 2 a ts: W 1 (a; 0), W 2 (a; 0). 7 aph. (a) 9 x 2 25 y 2 225 (b) 16 x 2 9 y 2](https://reader035.fdocuments.us/reader035/viewer/2022062415/5fdb545c9fe12d2da100204e/html5/thumbnails/34.jpg)
M-1
04
GE
NE
RA
LM
AT
HE
MA
TIC
S2
CH
AP
TE
R5:
AP
PL
ICA
TIO
NS
OF
INT
EG
RA
TIO
ND
ate
:/
/D
ay:
Page:
32
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
Exam
ple
70
Ske
tch
the
regi
on
byth
egr
aph
ofy
=x2,
then
fin
dit
sare
a.
Exam
ple
71
Ske
tch
the
regi
on
byth
egr
aphs
ofy
=x3
an
dy
=x
,th
enfi
nd
its
are
a.
Exam
ple
72
Ske
tch
the
regi
on
byth
egr
aphs
ofx
=√y
fromy
=0
an
dy
=1,
then
fin
dit
sare
a.
Exam
ple
73
Ske
tch
the
regi
on
byth
egr
aphs
ofy
=−x
+6,y
=√x
an
dy
=0,
then
fin
dit
sare
a.
![Page 35: (C) - KSUfac.ksu.edu.sa/sites/default/files/mdhkr_qdym_lmqrr_104...c) ertices: V (0 b), V 2 (0; b) 2 b. 2 a ts: W 1 (a; 0), W 2 (a; 0). 7 aph. (a) 9 x 2 25 y 2 225 (b) 16 x 2 9 y 2](https://reader035.fdocuments.us/reader035/viewer/2022062415/5fdb545c9fe12d2da100204e/html5/thumbnails/35.jpg)
M-1
04
GE
NE
RA
LM
AT
HE
MA
TIC
S2
CH
AP
TE
R5:
AP
PL
ICA
TIO
NS
OF
INT
EG
RA
TIO
ND
ate
:/
/D
ay:
Page:
33
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==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
(2)
Volu
mes
of
Soli
dof
Revolu
tion
Inth
isse
ctio
n,w
ew
ills
tart
look
ing
atth
evol
um
eofa
soli
dofre
volu
tion
.W
esh
ould
firs
tdefi
ne
just
wh
ata
soli
dof
revolu
tion
is.
Soli
dof
Revolu
tion
Defi
nit
ion
8T
he
soli
dof
revo
luti
on
(S)
isa
soli
dge
ner
ate
dfr
om
rota
tin
ga
regi
onR
abo
ut
ali
ne
inth
esa
me
pla
ne
wher
eth
eli
ne
isca
lled
the
axi
sof
revo
luti
on
.
Exam
ple
74
Letf
(x)≥
0be
con
tin
uou
sfo
rev
eryx∈
[a,b
].L
etR
bea
regi
on
bou
nded
byth
egr
aph
off
an
dx-
axi
sfo
rmx
=a
tox
=b.
Rota
tin
gth
ere
gionR
abo
ut
x-axi
sge
ner
ate
sa
soli
dgi
ven
inF
igu
re74.
Exam
ple
75
Letf
(x)
bea
con
stan
tfu
nct
ion
,as
inF
igu
re75.
The
regi
onR
isa
rect
an
gle
an
dro
tati
ng
itabo
ut
x-axi
sge
ner
ate
sa
circ
ula
rcy
lin
der
.
Exam
ple
76
Con
sider
the
regi
onR
bou
nded
byth
egr
aph
off
(y)
from
y=c
toy
=d
as
inF
igu
re.
Rev
olu
tion
ofR
abo
ut
y-axi
sge
ner
ate
sa
soli
dof
revo
luti
on
.
Volu
mes
of
Soli
dof
Revolu
tion
On
eof
the
sim
ple
stap
pli
cati
on
sof
inte
gra
tion
isto
det
erm
ine
avo
lum
eof
solid
of
revolu
tion
.In
this
sect
ion
,w
ew
ill
stu
dy
thre
em
eth
od
sto
evalu
ate
volu
mes
of
revolu
tion
kn
own
as
the
dis
km
eth
od
,th
ew
ash
erm
eth
od
an
dth
em
eth
od
of
cyli
nd
rica
lsh
ells
.
(A)
Dis
kM
eth
od
Letf
be
conti
nu
ou
son
[a,b
]an
dle
tR
be
are
gio
nb
ou
nd
edby
the
gra
ph
s,x-a
xis
an
dth
ep
oin
tsx
=a,x
=b.
LetS
be
aso
lid
gen
erate
dby
revo
lvin
gR
ab
ou
tx-a
xis
.A
ssu
meP
isa
part
itio
nof
[a,b
]an
dwk∈
[xk−1,xk]
isa
mark
er.
For
each
[xk−1,xk],
we
form
are
ctan
gle
,it
sh
igh
isf
(wk)
an
dit
sw
idth
is∆xk.
Th
ere
volu
tion
of
the
rect
an
gle
ab
ou
tx-a
xis
gen
erate
sa
circ
ula
rd
isk
.It
sra
diu
san
dh
igh
are
r=f
(wk),
h=
∆xk.
Th
evolu
me
of
each
circ
ula
rd
isk
is
Vk
=π
(f(w
k))
2∆xk.
Th
esu
mof
volu
mes
of
the
circ
ula
rd
isks
ap
pro
xim
ate
lygiv
esth
evo
lum
eof
the
soli
dof
revo
luti
on
:
V=
n ∑ k=1
∆Vk
=li
mn→∞
n ∑ k=1
π(f
(wk))
2∆xk
=π
∫ b a
[ f(x)] 2 d
x.
V=π
∫ b a
[ f(x)] 2 d
x.
![Page 36: (C) - KSUfac.ksu.edu.sa/sites/default/files/mdhkr_qdym_lmqrr_104...c) ertices: V (0 b), V 2 (0; b) 2 b. 2 a ts: W 1 (a; 0), W 2 (a; 0). 7 aph. (a) 9 x 2 25 y 2 225 (b) 16 x 2 9 y 2](https://reader035.fdocuments.us/reader035/viewer/2022062415/5fdb545c9fe12d2da100204e/html5/thumbnails/36.jpg)
M-1
04
GE
NE
RA
LM
AT
HE
MA
TIC
S2
CH
AP
TE
R5:
AP
PL
ICA
TIO
NS
OF
INT
EG
RA
TIO
ND
ate
:/
/D
ay:
Page:
34
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
Sim
ilar
ly,
we
fin
dth
evo
lum
eof
the
soli
dof
revo
luti
onab
ou
ty-a
xis
.L
etf
be
conti
nu
ous
on[c,d
]an
dle
tR
be
are
gion
bou
nd
edby
the
gra
ph
s,y-
axis
and
the
poi
ntsy
=c,y
=d.
LetS
be
aso
lid
gen
erate
dby
revo
lvin
gR
abou
ty-a
xis
.A
ssu
meP
isa
par
titi
onof
[c,d
]an
dwk∈
[yk−1,yk].
For
each
[yk−1,yk],
we
form
are
ctan
gle,
its
hig
hisf
(wk)
an
dit
sw
idth
is∆y k
.R
evol
uti
onof
each
rect
angl
eab
out
y-a
xis
gen
erate
sa
circ
ula
rd
isk
as
show
nin
.It
sra
diu
san
dh
igh
are
r=f
(wk),
h=
∆y k
.
Th
evo
lum
eof
the
soli
dof
revol
uti
ongi
ven
inis
the
sum
of
volu
mes
of
circ
ula
rd
isks
give
s:
V=
n ∑ k=1
∆Vk
=li
mn→∞
n ∑ k=1
π(f
(wk))
2∆y k
=π
∫ d c
[ f(y)] 2 d
y.
V=π
∫ d c
[ f(y)] 2 d
y.
Exam
ple
77
Ske
tch
the
regi
onR
bou
nded
byth
egr
aphs
of
the
equ
a-
tion
sy
=√x
,x
=4,y
=0.
Then
,fi
nd
the
volu
me
of
the
soli
dge
ner
ate
difR
isre
volv
edabo
ut
x-axi
s.
Exam
ple
78
Ske
tch
the
regi
onR
bou
nded
byth
egr
aphs
of
the
equ
a-
tion
sy
=ex
,y
=e
an
dx
=0.
Then
,fi
nd
the
volu
me
of
the
soli
dge
ner
ate
difR
isre
volv
edabo
ut
y-axi
s.
Exam
ple
79
Letx
=y2
on
the
inte
rval[0,1
].R
ota
teth
ere
gion
aro
un
dth
ey-
axi
san
dfi
nd
the
volu
me
of
the
resu
ltin
gso
lid.
![Page 37: (C) - KSUfac.ksu.edu.sa/sites/default/files/mdhkr_qdym_lmqrr_104...c) ertices: V (0 b), V 2 (0; b) 2 b. 2 a ts: W 1 (a; 0), W 2 (a; 0). 7 aph. (a) 9 x 2 25 y 2 225 (b) 16 x 2 9 y 2](https://reader035.fdocuments.us/reader035/viewer/2022062415/5fdb545c9fe12d2da100204e/html5/thumbnails/37.jpg)
M-1
04
GE
NE
RA
LM
AT
HE
MA
TIC
S2
CH
AP
TE
R5:
AP
PL
ICA
TIO
NS
OF
INT
EG
RA
TIO
ND
ate
:/
/D
ay:
Page:
35
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
(B)
Wash
er
Meth
od
Th
ew
ash
erm
eth
od
isth
ege
ner
aliz
atio
nof
the
dis
km
eth
od
for
are
gio
nb
etw
een
two
funct
ion
sf
(x)
andg(x
).L
etR
be
are
gio
nb
ou
nd
edby
the
grap
hs
off
(x)
andg(x
)su
chth
atf
(x)>g(x
)fr
omx
=a
to,x
=b.
Th
evo
lum
eof
the
soli
dS
gen
erat
edfr
omro
tati
ng
the
are
ab
ou
nd
edby
the
grap
hs
off
(x)
andg(x
)ar
oun
dx-a
xis
is
V=
∫ b a
[f(x
)]2dx−∫ b a
[g(x
)]2dx,
V=
∫ b a
( [f(x
)]2−
[g(x
)]2) dx
.
Sim
ilar
ly,
letR
be
are
gion
bou
nd
edby
the
gra
ph
soff
(y)
an
dg(y
)su
chth
atf
(y)>g(y
)fr
omy
=c
to,y
=d.
Th
evo
lum
eof
the
soli
dS
gen
erat
edfr
omro
tati
ng
the
area
bou
nd
edby
the
gra
ph
soff
an
dg
arou
nd
y-a
xis
is
V=
∫ d c
[f(y
)]2dy−∫ d c
[g(y
)]2dy,
V=
∫ d c
( [f(y
)]2−
[g(y
)]2) dy
.
Exam
ple
80
Eva
luate
the
volu
me
of
the
soli
dge
ner
ate
dby
revo
luti
on
of
the
bou
nded
regi
on
bygr
aphs
of
the
foll
ow
ing
two
fun
ctio
nsy
=x2
an
dy
=2x
abo
ut
x-axi
s.
![Page 38: (C) - KSUfac.ksu.edu.sa/sites/default/files/mdhkr_qdym_lmqrr_104...c) ertices: V (0 b), V 2 (0; b) 2 b. 2 a ts: W 1 (a; 0), W 2 (a; 0). 7 aph. (a) 9 x 2 25 y 2 225 (b) 16 x 2 9 y 2](https://reader035.fdocuments.us/reader035/viewer/2022062415/5fdb545c9fe12d2da100204e/html5/thumbnails/38.jpg)
M-1
04
GE
NE
RA
LM
AT
HE
MA
TIC
S2
CH
AP
TE
R5:
AP
PL
ICA
TIO
NS
OF
INT
EG
RA
TIO
ND
ate
:/
/D
ay:
Page:
36
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
Exam
ple
81
Con
sider
are
gionR
bou
nded
byth
egr
aphsy
=√x
,y
=6−x
an
dx-
axi
s.R
ota
teth
isre
gion
abo
ut
y-axi
san
dfi
nd
the
volu
me
of
the
gen
erate
dso
lid.
Exam
ple
82
Rec
on
sider
the
sam
ere
gion
as
inE
xam
ple
81
encl
ose
dby
the
curv
esy
=√x
,y
=6−x
an
dx-
axi
s.N
ow
rota
teth
isre
gion
abo
ut
the
x-axi
sin
stea
dan
dfi
nd
the
resu
ltin
gvo
lum
e.
![Page 39: (C) - KSUfac.ksu.edu.sa/sites/default/files/mdhkr_qdym_lmqrr_104...c) ertices: V (0 b), V 2 (0; b) 2 b. 2 a ts: W 1 (a; 0), W 2 (a; 0). 7 aph. (a) 9 x 2 25 y 2 225 (b) 16 x 2 9 y 2](https://reader035.fdocuments.us/reader035/viewer/2022062415/5fdb545c9fe12d2da100204e/html5/thumbnails/39.jpg)
M-1
04
GE
NE
RA
LM
AT
HE
MA
TIC
S2
CH
AP
TE
R5:
AP
PL
ICA
TIO
NS
OF
INT
EG
RA
TIO
ND
ate
:/
/D
ay:
Page:
37
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
(C)
Meth
od
of
Cyli
nd
rical
Sh
ell
s
Th
em
eth
od
ofcy
lin
dri
cal
shel
lsso
met
imes
easi
erth
an
the
wash
erm
eth
od
.T
his
isb
ecau
seso
lvin
geq
uat
ion
sfo
ron
eva
riab
lein
term
sof
anot
her
isn
otso
met
imes
sim
ple
(i.
e.,
solv
ingx
inte
rms
ofy
an
dve
rsa
vis
a).
For
exam
ple
,th
evol
um
eof
the
soli
dob
tain
edby
rota
tin
gth
ere
gion
bou
nd
edbyy
=2x
2−x3
andy
=0
ab
ou
tth
ey-a
xis
.B
yth
ew
ash
erm
eth
od
,w
ew
ould
hav
eto
solv
eth
ecu
bic
equ
ati
on
forx
inte
rms
ofy
and
this
isn
otsi
mp
le.
Inth
ew
ash
erm
eth
od
,w
eas
sum
eth
atth
ere
ctan
gle
from
each
sub
-in
terv
alis
vert
ical
toth
eax
isof
the
revo
luti
on
,b
ut
inth
em
eth
od
of
cyli
nd
rica
lsh
ells
,th
ere
ctan
gle
isp
aral
lel
toth
eaxis
of
revo
luti
on
.
As
show
nin
the
nex
tfi
gure
,le
tr 1
be
the
inn
erra
diu
sof
the
shel
l,r 2
be
the
oute
rra
diu
sof
the
shel
l,h
be
hig
hof
the
shel
l,∆r
=r 2−r 1
be
the
thic
kn
ess
ofth
esh
ell,
r=
r1+r2
2b
eth
eav
erag
era
diu
sof
the
shel
l.
Th
evol
um
eof
the
cylin
dri
cal
shel
lis
V=πr2 2h−πr2 1h
=π
(r2 2−r2 1
)h
=π
(r2
+r 1
)(r 2−r 1
)h
=2π
(r 2
+r 1
2)h
(r2−r 1
)
=2πrh
∆r.
Now
,co
nsi
der
the
grap
hgi
ven
inth
efi
gure
bel
ow.
Th
ere
volu
tion
of
the
regi
onR
abou
ty-a
xis
gen
erat
esa
soli
dgiv
enin
the
sam
efi
gu
re.
LetP
be
ap
arti
tion
ofth
ein
terv
al[a,b
]an
dle
twk
be
the
mid
poin
tof
[xk−1,xk].
Th
ere
volu
tion
ofth
ere
ctan
gle
abou
ty-a
xis
gen
erate
sa
cyli
nd
rica
lsh
ell
wh
ere
the
hig
h=f
(wk),
the
aver
age
rad
ius
=wk
an
dth
eth
icknes
s=
∆xk
.
Hen
ce,
the
volu
me
of
the
cyli
nd
rica
lsh
ell
Vk
=2πwkf
(wk)∆xk.
To
evalu
ate
the
volu
me
of
the
wh
ole
soli
d,
we
sum
the
volu
me
of
all
cyli
nd
rica
lsh
ells
.T
his
mea
ns
V=
n ∑ k=1
Vk
=2π
n ∑ k=1
wkf
(wk)∆xk.
Fro
mR
iem
an
nsu
m
n ∑ k=1
wkf
(wk)∆xk
=
∫ b a
xf
(x)dx
an
dth
isim
pli
es
V=
2π
∫ b a
xf
(x)dx.
Sim
ilarl
y,if
the
revolu
tion
of
the
regio
nab
ou
tx-a
xis
,th
evo
lum
eof
the
soli
dof
revo
luti
on
is
V=
2π
∫ d c
yf
(y)dy.
![Page 40: (C) - KSUfac.ksu.edu.sa/sites/default/files/mdhkr_qdym_lmqrr_104...c) ertices: V (0 b), V 2 (0; b) 2 b. 2 a ts: W 1 (a; 0), W 2 (a; 0). 7 aph. (a) 9 x 2 25 y 2 225 (b) 16 x 2 9 y 2](https://reader035.fdocuments.us/reader035/viewer/2022062415/5fdb545c9fe12d2da100204e/html5/thumbnails/40.jpg)
M-1
04
GE
NE
RA
LM
AT
HE
MA
TIC
S2
CH
AP
TE
R5:
AP
PL
ICA
TIO
NS
OF
INT
EG
RA
TIO
ND
ate
:/
/D
ay:
Page:
38
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
Exam
ple
83
Ske
tch
the
regi
onR
bou
nded
byth
egr
aphs
of
the
equ
a-
tion
sy
=2x−x2
an
dx
=0.
Then
,by
the
met
hod
of
cyli
ndri
cal
shel
ls,
fin
dth
evo
lum
eof
the
soli
dge
ner
ate
difR
isre
volv
edabo
ut
y-axi
s.
Exam
ple
84
Ske
tch
the
regi
onR
bou
nded
byth
egr
aphs
of
the
equ
a-
tion
sx
=√y
an
dx
=2.
Then
,fi
nd
the
volu
me
of
the
soli
dge
ner
ate
difR
isre
volv
edabo
ut
x-axi
s.
![Page 41: (C) - KSUfac.ksu.edu.sa/sites/default/files/mdhkr_qdym_lmqrr_104...c) ertices: V (0 b), V 2 (0; b) 2 b. 2 a ts: W 1 (a; 0), W 2 (a; 0). 7 aph. (a) 9 x 2 25 y 2 225 (b) 16 x 2 9 y 2](https://reader035.fdocuments.us/reader035/viewer/2022062415/5fdb545c9fe12d2da100204e/html5/thumbnails/41.jpg)
M-1
04
GE
NE
RA
LM
AT
HE
MA
TIC
S2
CH
AP
TE
R5:
AP
PL
ICA
TIO
NS
OF
INT
EG
RA
TIO
ND
ate
:/
/D
ay:
Page:
39
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
Hom
ew
ork
(5):
Qu
est
ion
1:
Ske
tch
the
regi
onby
the
grap
hs
ofy
=√x
an
dy
=x2,
then
fin
dit
sar
ea.
Qu
est
ion
2:
Ske
tch
the
regi
onby
the
grap
hs
ofy
=si
nx
an
dy
=co
sx
onth
ein
terv
al[0,π 4
],th
enfi
nd
its
area
.
Qu
est
ion
3:
Ske
tch
the
regi
onby
the
grap
hs
ofx
=y
+2,x
=y2
inth
efi
rst
qu
adra
nt,
then
fin
dit
sar
ea.
Qu
est
ion
4:
Eva
luat
eth
evol
um
eof
the
soli
dgen
erate
dby
revo
luti
on
ofth
eb
oun
ded
regi
onby
grap
hs
ofth
efo
llow
ing
fun
ctio
nsy
=−x
+2,
x=
0an
dy
=0
abou
ty-a
xis
.
Qu
est
ion
5:
Eva
luate
the
volu
me
of
the
solid
gen
erate
dby
revolu
tion
of
the
bou
nd
edre
gio
nby
gra
ph
sof
the
foll
owin
gtw
ofu
nct
ion
sy
=x2
an
dy
=x
ab
ou
tx-a
xis
.
Qu
est
ion
6:
Eva
luate
the
volu
me
of
the
solid
gen
erate
dby
revolu
tion
of
the
bou
nd
edre
gio
nby
gra
ph
sof
the
foll
owin
gtw
ofu
nct
ion
sy
=x3
an
dy
=x
fromx
=0
tox
=1 2
ab
ou
tx-a
xis
.
![Page 42: (C) - KSUfac.ksu.edu.sa/sites/default/files/mdhkr_qdym_lmqrr_104...c) ertices: V (0 b), V 2 (0; b) 2 b. 2 a ts: W 1 (a; 0), W 2 (a; 0). 7 aph. (a) 9 x 2 25 y 2 225 (b) 16 x 2 9 y 2](https://reader035.fdocuments.us/reader035/viewer/2022062415/5fdb545c9fe12d2da100204e/html5/thumbnails/42.jpg)
M-1
04
GE
NE
RA
LM
AT
HE
MA
TIC
S2
CH
AP
TE
R6:
PA
RT
IAL
DE
RIV
AT
IVE
SD
ate
:/
/D
ay:
Page:
40
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
(1)
Fun
cti
on
sof
two
vari
ab
les:
Anti
-deri
vati
ves
Defi
nit
ion
9A
fun
ctio
nof
two
vari
abl
esis
aru
leth
at
ass
ign
san
ord
ered
pair
(x,y
)(i
nth
edom
ain
of
the
fun
ctio
n)
toa
real
nu
mbe
rw
.f
:R
2−→
R
(x,y
)−→
w
Exam
ple
85
1.f
(x,y
)=x2
+y2
2.f
(x,y
)=
xx2+y2
Part
ial
deri
vati
ves
of
afu
ncti
on
of
two
vari
ab
les
Ifw
=f
(x,y
)is
afu
nct
ion
oftw
ova
riab
les,
then
:1.
Th
ep
arti
ald
eriv
ativ
eoff
wit
hre
spec
ttox
isd
enote
dby∂f∂x,∂w∂x,f x
orwx,
and
itis
calc
ula
ted
by
app
lyin
gth
eru
les
of
diff
eren
tiati
on
tox
and
rega
rdin
gy
asa
con
stan
t.
2.T
he
par
tial
der
ivat
ive
off
wit
hre
spec
ttoy
isd
enote
dby∂f∂y,∂w∂y,f y
orwy,
and
itis
calc
ula
ted
by
app
lyin
gth
eru
les
of
diff
eren
tiati
on
toy
and
rega
rdin
gx
asa
con
stan
t.
Exam
ple
86
Calc
ula
tef x
an
df y
of
the
fun
ctio
nf
(x,y
)=
x2y3
+xyln
(x+y)
(2)
Fu
ncti
on
sof
thre
evari
ab
les
:
Anti
-deri
vati
ves
Defi
nit
ion
10
Afu
nct
ion
of
two
vari
abl
esis
aru
leth
at
ass
ign
san
ord
ered
pair
(x,y
)(i
nth
edom
ain
of
the
fun
ctio
n)
toa
real
nu
mbe
rw
.f
:R
3−→
R
(x,y,z
)−→
w
Exam
ple
87f
(x,y
)=x2
+y2
+z
Part
ial
deri
vati
ves
of
afu
ncti
on
of
thre
evari
ab
les
Ifw
=f
(x,y,z
)is
afu
nct
ion
of
thre
eva
riab
les,
then
:1.
Th
ep
art
iald
eriv
ati
ve
off
wit
hre
spec
ttox
isd
enote
dby∂f∂x,∂w∂x,f x
orwx,
an
dit
isca
lcu
late
dby
ap
ply
ing
the
rule
sof
diff
eren
tiati
on
tox
an
dre
gard
ingy
an
dz
as
aco
nst
ant.
2.
Th
ep
art
iald
eriv
ati
ve
off
wit
hre
spec
ttoy
isd
enote
dby∂f∂y,∂w∂y,f y
orwy,
an
dit
isca
lcu
late
dby
ap
ply
ing
the
rule
sof
diff
eren
tiati
on
toy
an
dre
gard
ingx
an
dz
as
aco
nst
ant.
3.
Th
ep
art
iald
eriv
ati
ve
off
wit
hre
spec
ttoz
isd
enote
dby∂f∂z,∂w∂z,f z
orwz,
an
dit
isca
lcu
late
dby
ap
ply
ing
the
rule
sof
diff
eren
tiati
on
toz
an
dre
gard
ingx
an
dy
as
aco
nst
ant.
Exam
ple
88
Calc
ula
tef x
,f y
an
df z
of
the
fun
ctio
nf
(x,y
)=z2y3−
y2(x
3+z)
![Page 43: (C) - KSUfac.ksu.edu.sa/sites/default/files/mdhkr_qdym_lmqrr_104...c) ertices: V (0 b), V 2 (0; b) 2 b. 2 a ts: W 1 (a; 0), W 2 (a; 0). 7 aph. (a) 9 x 2 25 y 2 225 (b) 16 x 2 9 y 2](https://reader035.fdocuments.us/reader035/viewer/2022062415/5fdb545c9fe12d2da100204e/html5/thumbnails/43.jpg)
M-1
04
GE
NE
RA
LM
AT
HE
MA
TIC
S2
CH
AP
TE
R6:
PA
RT
IAL
DE
RIV
AT
IVE
SD
ate
:/
/D
ay:
Page:
41
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
Secon
dp
art
ial
deri
vati
ves:
Ifw
=f
(x,y
)is
afu
nct
ion
oftw
ova
riab
les,
then
:
1.∂2f
∂x2
=∂ ∂x
(∂f∂x
)=
∂ ∂xf x
=f xx
2.∂2f
∂y2
=∂ ∂y(∂f∂y
)=
∂ ∂yf y
=f yy
3.∂2f
∂x∂y
=∂ ∂x
(∂f∂y
)=
∂ ∂xf y
=f xy
4.∂2f
∂y∂x
=∂ ∂y(∂f∂x
)=
∂ ∂yf x
=f yx
Note
:S
econ
dp
arti
ald
eriv
ativ
esof
afu
nct
ion
of
thre
eva
riable
sare
defi
ned
ina
sam
em
ann
er.
Anti
-deri
vati
ves
Th
eore
m5
Letf
(x,y
)be
afu
nct
ion
of
two
vari
abl
es.
Iff
,f x
,f y
,f xy
an
df yx
are
con
tin
uou
s,th
enf xy
=f yx.
Note
:Iff
(x,y,z
)is
afu
nct
ion
ofth
ree
vari
able
sandf
has
conti
nuou
sse
con
dp
arti
ald
eriv
ativ
es,
then
f xy
=f yx,fxz
=f zx
an
df yz
=f zy.
Exam
ple
89
Letf
(x,y
)=x3y
+xy2sin
(x+y)
,ca
lcu
latef xy
an
df yx.
Exam
ple
90
Letf
(x,y
)=z3x
+y2(x
+yz)
,ca
lcu
latef xy
an
df yx.
1.f x
,f y
an
df z
at
(1,1,1
).
2.f xx,f yy
an
df zz.
3.f xy,f yz
an
df zx
at
(0,−
1,1).
![Page 44: (C) - KSUfac.ksu.edu.sa/sites/default/files/mdhkr_qdym_lmqrr_104...c) ertices: V (0 b), V 2 (0; b) 2 b. 2 a ts: W 1 (a; 0), W 2 (a; 0). 7 aph. (a) 9 x 2 25 y 2 225 (b) 16 x 2 9 y 2](https://reader035.fdocuments.us/reader035/viewer/2022062415/5fdb545c9fe12d2da100204e/html5/thumbnails/44.jpg)
M-1
04
GE
NE
RA
LM
AT
HE
MA
TIC
S2
CH
AP
TE
R6:
PA
RT
IAL
DE
RIV
AT
IVE
SD
ate
:/
/D
ay:
Page:
42
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==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
Ch
ain
Ru
les
1.Ifw
=f
(x,y
)an
dx
=g(t
),y
=h
(t),
such
thatf
,g
an
dh
are
diff
eren
tiab
leth
en
df dt
=dw dt
=∂w∂x
dx dt
+∂w∂y
dydt
2.Ifw
=f
(x,y
)an
dx
=g(t,s
),y
=h
(t,s
),su
chth
atf
,g
an
dh
are
diff
eren
tiab
leth
en
∂f∂t
=∂w ∂t
=∂w∂x
∂x∂t
+∂w∂y
∂y∂t
∂f∂s
=∂w∂s
=∂w∂x
∂x∂s
+∂w∂y
∂y∂s
3.Ifw
=f
(x,y,z
)an
dx
=g(t,s
),y
=h
(t,s
),z
=k(t,s
),su
chth
atf
,g,h
andk
are
diff
eren
tiab
leth
en
∂f∂t
=∂w ∂t
=∂w∂x
∂x∂t
+∂w∂y
∂y∂t
+∂w∂z
∂z∂t
∂f∂s
=∂w∂s
=∂w∂x
∂x∂s
+∂w∂y
∂y∂s
+∂w∂z
∂z∂s
Exam
ple
91
Letf
(x,y
)=xy
+y2,x
=s2t,
an
dy
=s
+t
,ca
lcu
late
1.∂f∂t
2.∂f∂s
Exam
ple
92
Letf
(x,y,z
)=x
+sin
(xy)
+cos(xz),x
=ts,y
=s
+t
an
dz
=s t,
calc
ula
te
1.∂f∂t
2.∂f∂s
![Page 45: (C) - KSUfac.ksu.edu.sa/sites/default/files/mdhkr_qdym_lmqrr_104...c) ertices: V (0 b), V 2 (0; b) 2 b. 2 a ts: W 1 (a; 0), W 2 (a; 0). 7 aph. (a) 9 x 2 25 y 2 225 (b) 16 x 2 9 y 2](https://reader035.fdocuments.us/reader035/viewer/2022062415/5fdb545c9fe12d2da100204e/html5/thumbnails/45.jpg)
M-1
04
GE
NE
RA
LM
AT
HE
MA
TIC
S2
CH
AP
TE
R6:
PA
RT
IAL
DE
RIV
AT
IVE
SD
ate
:/
/D
ay:
Page:
43
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
Imp
licit
diff
ere
nti
ati
on
1.S
up
pos
eth
atth
eeq
uat
ionF
(x,y
)=
0d
efin
esy
imp
lici
tly
as
afu
nct
ion
ofx
sayy
=f
(x),
then dy
dx
=−Fx
Fy
2.S
up
pos
eth
atth
eeq
uat
ionF
(x,y,z
)=
0im
pli
citl
yd
efin
esa
fun
ctio
nz
=f
(x,y
),w
her
ef
isd
iffer
enti
able
,th
en
∂z
∂x
=−Fx
Fz
and∂z
∂x
=−Fy
Fz
Exam
ple
93
Lety2−xy
+3x2
=0,
fin
ddydx
.
Exam
ple
94
LetF
(x,y,z
)=x2y
+z2
+sin
(xyz)
=0
,fi
nd
∂z∂x
an
d∂z∂y
.
![Page 46: (C) - KSUfac.ksu.edu.sa/sites/default/files/mdhkr_qdym_lmqrr_104...c) ertices: V (0 b), V 2 (0; b) 2 b. 2 a ts: W 1 (a; 0), W 2 (a; 0). 7 aph. (a) 9 x 2 25 y 2 225 (b) 16 x 2 9 y 2](https://reader035.fdocuments.us/reader035/viewer/2022062415/5fdb545c9fe12d2da100204e/html5/thumbnails/46.jpg)
M-1
04
GE
NE
RA
LM
AT
HE
MA
TIC
S2
CH
AP
TE
R6:
PA
RT
IAL
DE
RIV
AT
IVE
SD
ate
:/
/D
ay:
Page:
44
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
Hom
ew
ork
(6):
Qu
est
ion
1:
Iff
(x,y
)=
2x4y3−xy2
+3y
+1,
fin
d1)f x
2)f y
3)f xy
4)f xx
Qu
est
ion
2:
Iff
(x,y
)=
4ex2y3
,
fin
d1)f x
2)f y
3)f xy
4)f xx
5)f yy
at(1,1
).
Qu
est
ion
3:
Ifx
=2s
+t,y
=s
lnt,w
=x2
+2xy
at
(1,1
)fi
nd
1)∂w∂s
2)∂w∂s
at(1,1
).
Qu
est
ion
4:
By
usi
ng
the
imp
lici
tfu
nct
ion
diff
eren
tiati
on
,fi
nddydx
1)x3−
3xy2
+y3
=5
2)x−√xy
+3y
=4
3)
2x3
+x2y
+y3
=1
![Page 47: (C) - KSUfac.ksu.edu.sa/sites/default/files/mdhkr_qdym_lmqrr_104...c) ertices: V (0 b), V 2 (0; b) 2 b. 2 a ts: W 1 (a; 0), W 2 (a; 0). 7 aph. (a) 9 x 2 25 y 2 225 (b) 16 x 2 9 y 2](https://reader035.fdocuments.us/reader035/viewer/2022062415/5fdb545c9fe12d2da100204e/html5/thumbnails/47.jpg)
M-1
04
GE
NE
RA
LM
AT
HE
MA
TIC
S2
CH
AP
TE
R7:
DIF
FE
RE
NT
IAL
EQ
UA
TIO
NS
Date
:/
/D
ay:
Page:
45
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==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
Defi
nit
ion
of
ad
iffere
nti
al
equ
ati
on
Defi
nit
ion
11
An
equ
ati
on
that
invo
lvesx,y,y′ ,y′′,y′′′,y
(4),...,y
(n)
for
afu
nc-
tiony(x
)w
ithnth
der
ivati
vey(n
)ofy
wit
hre
spec
ttox
isan
ord
inary
diff
eren
tial
Equ
ati
on
of
ord
ern
.
Exam
ple
95
1.y′
=x2
+5
isa
diff
eren
tial
equ
ati
on
of
ord
er1.
2.y′′
+x
(y′ )3−y
=x
isa
diff
eren
tial
equ
ati
on
of
ord
er2.
3.
(y(4
))3
+x2y′′
=2x
isa
diff
eren
tial
equ
ati
on
of
ord
er4.
Rem
ark
3y
=y(x
)is
call
eda
solu
tion
of
adiff
eren
tial
equ
ati
on
ify
=y(x
)sa
tisfi
esth
at
diff
eren
tial
equ
ati
on
.
Exam
ple
96
Con
sider
the
diff
eren
tial
equ
ati
ony′
=6x
+4,
then
y=
3x2
+4x
isa
solu
tion
of
that
diff
eren
tial
equ
ati
on
.
Note
:1.y
=3x
2+
4x
isth
ege
ner
also
luti
onof
that
diff
eren
tial
equ
ati
on
.
2.If
anin
itia
lco
nd
itio
nw
asad
ded
toth
ed
iffer
enti
al
equ
ati
on
toass
ign
ace
rtai
nva
lue
forc
then
y=y(x
)is
call
edth
ep
art
icu
lar
solu
tion
of
the
diff
eren
tial
equ
atio
n.
Exam
ple
97
Con
sider
the
diff
eren
tial
equ
ati
ony′
=6x
+4
wit
hth
ein
itia
lco
ndit
iony(0
)=
2,y
=3x2
+4x
+c
isth
ege
ner
al
solu
tion
of
the
diff
eren
tial
equ
ati
on
,
y(0
)=
2⇒
3(0)
2+
4(0)
+c
=2⇒c
=2.
Hen
cey
=3x2
+4x
+2
isth
epa
rtic
ula
rso
luti
on
of
the
diff
eren
tial
equ
ati
on
.
Sep
ara
ble
Diff
ere
nti
al
equ
ati
on
s
Th
ese
para
ble
diff
eren
tial
equ
ati
on
has
the
form
M(x
)+N
(y)y′
=0
wh
ereM
(x)
an
dN
(y)
are
conti
nu
ou
sfu
nct
ion
s.T
oso
lve
the
sep
ara
ble
diff
eren
tial
equ
ati
on
:1.
Wri
teth
eeq
uati
on
asM
(x)dx
+N
(y)dy
=0⇒N
(y)dy
=−M
(x)dx
. 2.
Inte
gra
teth
ele
ft-h
an
dsi
de
wit
hre
spec
ttoy
an
dth
eri
ght-
han
dsi
de
wit
hre
spec
ttox
.
Exam
ple
98
Solv
eth
ediff
eren
tial
equ
ati
ony′+y3ex
=0
.
![Page 48: (C) - KSUfac.ksu.edu.sa/sites/default/files/mdhkr_qdym_lmqrr_104...c) ertices: V (0 b), V 2 (0; b) 2 b. 2 a ts: W 1 (a; 0), W 2 (a; 0). 7 aph. (a) 9 x 2 25 y 2 225 (b) 16 x 2 9 y 2](https://reader035.fdocuments.us/reader035/viewer/2022062415/5fdb545c9fe12d2da100204e/html5/thumbnails/48.jpg)
M-1
04
GE
NE
RA
LM
AT
HE
MA
TIC
S2
CH
AP
TE
R7:
DIF
FE
RE
NT
IAL
EQ
UA
TIO
NS
Date
:/
/D
ay:
Page:
46
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
Exam
ple
99
Solv
eth
ediff
eren
tial
equ
ati
ondydx
=y2ex
,y(0
)=
1
Exam
ple
100
Solv
eth
ediff
eren
tial
equ
ati
ondy−sinx
(1+y2)dx
=0
.
Exam
ple
101
Solv
eth
ediff
eren
tial
equ
ati
one−
ysinx−y′ cos
2x
=0
Exam
ple
102
Solv
eth
ediff
eren
tial
equ
ati
ony′
=1−y
+x2−yx2
.
![Page 49: (C) - KSUfac.ksu.edu.sa/sites/default/files/mdhkr_qdym_lmqrr_104...c) ertices: V (0 b), V 2 (0; b) 2 b. 2 a ts: W 1 (a; 0), W 2 (a; 0). 7 aph. (a) 9 x 2 25 y 2 225 (b) 16 x 2 9 y 2](https://reader035.fdocuments.us/reader035/viewer/2022062415/5fdb545c9fe12d2da100204e/html5/thumbnails/49.jpg)
M-1
04
GE
NE
RA
LM
AT
HE
MA
TIC
S2
CH
AP
TE
R7:
DIF
FE
RE
NT
IAL
EQ
UA
TIO
NS
Date
:/
/D
ay:
Page:
47
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
Th
efi
rst-
ord
erli
nea
rd
iffer
enti
aleq
uat
ion
has
the
form
y′+P
(x)y
=Q
(x),
wh
ereP
(x)
andQ
(x)
are
conti
nu
ous
fun
ctio
ns
ofx
To
solv
eth
efi
rst-
ord
erli
nea
rd
iffer
enti
aleq
uati
on
:1.
Com
pute
the
inte
grat
ing
fact
oru
(x)
=e∫ P(
x)dx
.
2.T
he
gen
eral
solu
tion
ofth
efi
rst-
ord
erli
nea
rd
iffer
enti
al
equ
ati
on
is
y(x
)=
1
u(x
)
∫ u(x
)Q(x
)dx
Exam
ple
103
Solv
eth
ediff
eren
tial
equ
ati
onxdydx
+y
=x2
+1
.
Exam
ple
104
Solv
eth
ediff
eren
tial
equ
ati
ony′−
2 xy
=x2ex
,y(1
)=e
.
![Page 50: (C) - KSUfac.ksu.edu.sa/sites/default/files/mdhkr_qdym_lmqrr_104...c) ertices: V (0 b), V 2 (0; b) 2 b. 2 a ts: W 1 (a; 0), W 2 (a; 0). 7 aph. (a) 9 x 2 25 y 2 225 (b) 16 x 2 9 y 2](https://reader035.fdocuments.us/reader035/viewer/2022062415/5fdb545c9fe12d2da100204e/html5/thumbnails/50.jpg)
M-1
04
GE
NE
RA
LM
AT
HE
MA
TIC
S2
CH
AP
TE
R7:
DIF
FE
RE
NT
IAL
EQ
UA
TIO
NS
Date
:/
/D
ay:
Page:
48
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
Exam
ple
105
Solv
eth
ediff
eren
tial
equ
ati
ony′+y
=cos(ex
).
Exam
ple
106
Solv
eth
ediff
eren
tial
equ
ati
onxy′−
3y=x2
.
![Page 51: (C) - KSUfac.ksu.edu.sa/sites/default/files/mdhkr_qdym_lmqrr_104...c) ertices: V (0 b), V 2 (0; b) 2 b. 2 a ts: W 1 (a; 0), W 2 (a; 0). 7 aph. (a) 9 x 2 25 y 2 225 (b) 16 x 2 9 y 2](https://reader035.fdocuments.us/reader035/viewer/2022062415/5fdb545c9fe12d2da100204e/html5/thumbnails/51.jpg)
M-1
04
GE
NE
RA
LM
AT
HE
MA
TIC
S2
CH
AP
TE
R7:
DIF
FE
RE
NT
IAL
EQ
UA
TIO
NS
Date
:/
/D
ay:
Page:
49
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
Hom
ew
ork
(7):
Qu
est
ion
1:
Sol
veth
ed
iffer
enti
aleq
uat
ion
s:
1)x2dy
+y2dx
=0
2)co
s2xdy−y2dx
=0
3)xdydx−
2y=x3secx
tanx
4)y′
=x2
+y2
5)y′+
3y
=e−
2x
Qu
est
ion
2:
Solv
eth
ed
iffer
enti
al
equ
ati
ony′+
2y
=x
,y(0
)=
1
Qu
est
ion
3:
Solv
eth
ed
iffer
enti
al
equ
ati
onxy′+y
=si
nx
,y(π 3
)=
2
Qu
est
ion
4:
Solv
eth
ed
iffer
enti
al
equ
ati
ondydx
+y−
1ex+1
=0
![Page 52: (C) - KSUfac.ksu.edu.sa/sites/default/files/mdhkr_qdym_lmqrr_104...c) ertices: V (0 b), V 2 (0; b) 2 b. 2 a ts: W 1 (a; 0), W 2 (a; 0). 7 aph. (a) 9 x 2 25 y 2 225 (b) 16 x 2 9 y 2](https://reader035.fdocuments.us/reader035/viewer/2022062415/5fdb545c9fe12d2da100204e/html5/thumbnails/52.jpg)
M-1
04
GE
NE
RA
LM
AT
HE
MA
TIC
S2
CH
AP
TE
R8:
PO
LA
RC
OO
RD
INA
TE
San
dA
PP
LIC
AT
ION
SD
ate
:/
/D
ay:
Page:
50 ==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
(1)
Defi
nit
ion
:
Defi
nit
ion
12
The
pola
rco
ord
inate
syst
emis
atw
o-
dim
ensi
on
al
coord
inate
syst
emin
whic
hea
chpo
intP
on
apla
ne
isdet
erm
ined
bya
dis
tan
cer
from
afi
xed
poin
tO
that
isca
lled
the
pole
(or
ori
gin
)an
dan
an
gleθ
from
afi
xed
dir
ecti
on
.
Rem
ark
4In
the
pola
rco
ord
inate
s(r,θ
),ifr>
0,th
epo
int
(r,θ
)li
esin
the
sam
equ
adra
nt
asθ;
ifr<
0,it
lies
inth
equ
adra
nt
on
the
oppo
site
side
of
the
pole
.M
ean
ing
that,
the
pola
rco
ord
inate
s(r,θ
)an
d(−r,θ)
lie
inth
esa
me
lin
eth
rou
ghth
epo
leO
an
dat
the
sam
edis
tan
ce|r|f
romO
,bu
ton
oppo
site
sides
ofO
.
Exam
ple
107
Plo
tth
epo
ints
whose
pola
rco
ord
inate
sare
give
n:
1.
(1,π/4
)
2.
(2,3π
)
3.
(2,−π/3
)
4.
(−3,π/6
)
(2)
Rela
tion
ship
betw
een
Recta
ngu
lar
an
dP
ola
rC
oord
inate
sL
et(x,y
)b
ea
rect
an
gu
lar
coord
inate
an
d(r,θ
)b
ea
pola
rco
ord
inate
.L
etth
ep
ole
at
the
ori
gin
poin
tan
dp
ola
raxis
on
x-a
xis
,an
dth
eli
ne
θ=
π 2on
y-a
xis
as
show
nin
Fig
ure
.
Fro
mth
etr
ian
gle
OA
P
cosθ
=x r⇒x
=r
cosθ,
sinθ
=y r⇒y
=r
sinθ.
Thu
s,
x2
+y2
=(r
cosθ)
2+
(rsi
nθ)
2,
=r2
(cos2θ
+si
n2θ)
x2
+y2
=r2.
x=
rco
sθ,
y=
rsi
nθ
tanθ
=y x,x2
+y2
=r2
Exam
ple
108
Con
vert
the
poin
tsfr
om
the
pola
rco
ord
inate
sto
the
rect
-an
gula
rco
ord
inate
s:1.
(1,π/4
)2.
(2,π
)
![Page 53: (C) - KSUfac.ksu.edu.sa/sites/default/files/mdhkr_qdym_lmqrr_104...c) ertices: V (0 b), V 2 (0; b) 2 b. 2 a ts: W 1 (a; 0), W 2 (a; 0). 7 aph. (a) 9 x 2 25 y 2 225 (b) 16 x 2 9 y 2](https://reader035.fdocuments.us/reader035/viewer/2022062415/5fdb545c9fe12d2da100204e/html5/thumbnails/53.jpg)
M-1
04
GE
NE
RA
LM
AT
HE
MA
TIC
S2
CH
AP
TE
R8:
PO
LA
RC
OO
RD
INA
TE
San
dA
PP
LIC
AT
ION
SD
ate
:/
/D
ay:
Page:
51 ==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
Exam
ple
109
Con
vert
the
poin
tsfr
om
the
rect
an
gula
rco
ord
inate
sto
pola
rco
ord
inate
s:1.
(5,0
)2.
(2√
3,2)
(3)
Sketc
hof
Pola
rC
urv
es
Cir
cle
sin
pola
rcoord
inate
s1.
Aci
rcle
its
cente
ratO
and
rad
iusa:r
=a
.2.
Aci
rcle
its
cente
rat
(a,0
)an
dra
diu
s|a|:r
=2a
cosθ
.3.
Aci
rcle
its
cente
rat
(0,a
)an
dra
diu
s|a|:r
=2a
sinθ
.
(4)
Are
ain
Pola
rC
oord
inate
sL
etr
=f
(θ)
be
aco
nti
nu
ous
fun
ctio
non
the
inte
rval
[α,β
]su
chth
at
0≤α≤β≤
2π.
Letf
(θ)>
0ov
erth
atin
terv
al
an
dR
be
ap
ola
rre
gion
bou
nd
edby
the
pol
areq
uat
ion
sr
=f
(θ),θ
=α
an
dθ
=β
as
show
nin
Fig
ure
.
A=
1 2
∫ β α
( f(θ)) 2 d
θ
Sim
ilarl
y,ass
um
ef
an
dg
are
conti
nu
ou
son
the
inte
rval
[α,β
]su
chth
at
f(θ
)>g(θ
).T
he
are
ab
ou
nd
edby
the
curv
esoff
an
dg
on
the
inte
rval
[α,β
]is
A=
1 2
∫ β α
[( f(θ)) 2 −
( g(θ)) 2] d
θ
Exam
ple
110
Fin
dth
eare
aof
the
regi
on
bou
nded
byth
egr
aph
of
the
pola
req
uati
onr
=2.
Exam
ple
111
Fin
dth
eare
aof
regi
on
lyin
gin
the
firs
tqu
adra
nt
an
din
side
the
circ
lew
ith
pola
req
uati
onr
=2.
Exam
ple
112
Fin
dth
eare
aof
regi
on
lyin
gin
side
the
circ
lew
ith
pola
req
uati
onr
=2
an
dou
tsid
eth
eci
rcle
wit
hpo
lar
equ
ati
onr
=1.
![Page 54: (C) - KSUfac.ksu.edu.sa/sites/default/files/mdhkr_qdym_lmqrr_104...c) ertices: V (0 b), V 2 (0; b) 2 b. 2 a ts: W 1 (a; 0), W 2 (a; 0). 7 aph. (a) 9 x 2 25 y 2 225 (b) 16 x 2 9 y 2](https://reader035.fdocuments.us/reader035/viewer/2022062415/5fdb545c9fe12d2da100204e/html5/thumbnails/54.jpg)
M-1
04
GE
NE
RA
LM
AT
HE
MA
TIC
S2
CH
AP
TE
R8:
PO
LA
RC
OO
RD
INA
TE
San
dA
PP
LIC
AT
ION
SD
ate
:/
/D
ay:
Page:
52 ==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
Hom
ew
ork
(8):
Qu
est
ion
1:
Con
vert
the
poi
nts
from
the
pola
rco
ord
inate
sto
the
rect
angu
lar
coor
din
ates
:1.
(3,π/2
)
2.(2,π 3
)
Qu
est
ion
2:
Con
vert
the
poi
nts
from
the
rect
an
gu
lar
coord
inate
sto
the
pol
arco
ord
inat
es:
1.(4,0
)
2.(6,3
)
Qu
est
ion
3:
Fin
dth
eare
aof
regio
nly
ing
insi
de
the
circ
lew
ith
pola
req
uati
onr
=2.
Qu
est
ion
4:
Fin
dth
eare
aof
regio
nly
ing
inth
efi
rst
qu
ad
rant
an
din
sid
eth
eci
rcle
wit
hp
ola
req
uati
onr
=3.
Qu
est
ion
5:
Fin
dth
eare
aof
regio
nly
ing
insi
de
the
circ
lew
ith
pola
req
uati
onr
=1
an
dou
tsid
eth
eci
rcle
wit
hp
ola
req
uati
onr
=2.
Qu
est
ion
6:
Fin
dth
eare
aof
regio
nly
ing
inth
efi
rst
qu
ad
rant
an
din
sid
eth
eci
rcle
wit
hp
ola
req
uati
onr
=2
an
dou
tsid
eth
eci
rcle
wit
hp
ola
req
uati
onr
=3.
![Page 55: (C) - KSUfac.ksu.edu.sa/sites/default/files/mdhkr_qdym_lmqrr_104...c) ertices: V (0 b), V 2 (0; b) 2 b. 2 a ts: W 1 (a; 0), W 2 (a; 0). 7 aph. (a) 9 x 2 25 y 2 225 (b) 16 x 2 9 y 2](https://reader035.fdocuments.us/reader035/viewer/2022062415/5fdb545c9fe12d2da100204e/html5/thumbnails/55.jpg)
M-1
04
GE
NE
RA
LM
AT
HE
MA
TIC
S2
An
swers
of
som
equ
est
ion
sP
age:
53
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
CH
AP
TE
R1:
1.F
(1,1
),D
:y
=−
3
2.(y−
2)2
=2(x
+4)
3.(x−
3)2
=8(y−
4)
4.x2
16
+y2 9
=1
5.(x−7)2
25
+(y
+2)2
16
=1
6.y2 4−
x2 9
=1
7.(x−7)2
1−
(y+2)2
8=
1
8.(x
+3)2
5+
(y−4)2
1=
1
9.(x
+1)
2=−
2(y−
3 2)
10.
(y+3)2
5−
(x+4)2
1=
1
CH
AP
TE
R2:
Q3:
1)11
2)73
CH
AP
TE
R3:
Qu
esti
on1:
1.X
=
5 6 7 2.X
=
6 8 10
Q
ues
tion
2:1.
X=
5 6 7
2.X
=
6 4 2
Q
ues
tion
3:1.
X=
7/2
−7/2 7
2.X
=
6 8 10
CH
AP
TE
R4:
Qu
esti
on
1:
1.
tan(3x−5)
3+c
2.
sin−1(x 4
)+c
3.xex−ex
+c
4.x
sinx
+co
sx
+c
5.xsin−1x
+√
1−x2
+c
6.
1 3ln|x−
2|−
1 3ln|x
+1|+
c
7.
(2x2−3)9
36
+c
8.
3si
n3√x
+c
Qu
esti
on
2:
1.
tanx
+secx
+c
2.x2(lnx−1)
8+c
3.
32 3
4.
ln3−
ln(e−
2)
5.
2ln
7−
ln4
6.−
31 2
CH
AP
TE
R5:
Q1.
5 12
Q2.
2−√2
√2
Q3.
27 6
Q4.
64π
3Q
5.
2π
15
CH
AP
TE
R6:
Q2:
1)
8e2)
12e
3)
48e
4)
24e
5)
60e
Q3:
1)
12
2)
12
Q4:
1)−
3x2−3y2
−6xy+3y2
2)−
2√xy−y
−x+6√xy
3)−
6x2+2xy
x2+3y2
CH
AP
TE
R7:
Q1:
1)y
=−x
1+cx
2)y
=−1
c+ta
nx
Q4:y
=1 ex
ln(ex
+1)
+c
CH
AP
TE
R8:
Q1:
1)
(0,3
)Q
2:
2)
(2,π 6
)Q
6:
5π 2