Math 152.02 Calculus with Analytic Geometry II€¦ · Lecture 6 - 1/14 Sketching antiderivatives...
Transcript of Math 152.02 Calculus with Analytic Geometry II€¦ · Lecture 6 - 1/14 Sketching antiderivatives...
Math 152.02
Lecture 4 - 1/10
Integrals andarea
TheFundamentalTheorem ofCalculus
Interpretingintegrals
Average value ofa function
Lecture 5 - 1/12
Properties ofdefinite integrals
Bounding onintegrals
Antiderivatives
Lecture 6 - 1/14
Sketchingantiderivatives
Basic integrationrules
Math 152.02Calculus with Analytic Geometry II
January 14, 2011
Math 152.02
Lecture 4 - 1/10
Integrals andarea
TheFundamentalTheorem ofCalculus
Interpretingintegrals
Average value ofa function
Lecture 5 - 1/12
Properties ofdefinite integrals
Bounding onintegrals
Antiderivatives
Lecture 6 - 1/14
Sketchingantiderivatives
Basic integrationrules
Integrals and area
Integrals and area
If a < b then∫ b
a
f (x) dx
= (area under f above x-axis)− (area above f under x-axis)
Example 18
∫ 0
−3
√9− x2 dx = π·32
4 = 9π4
Math 152.02
Lecture 4 - 1/10
Integrals andarea
TheFundamentalTheorem ofCalculus
Interpretingintegrals
Average value ofa function
Lecture 5 - 1/12
Properties ofdefinite integrals
Bounding onintegrals
Antiderivatives
Lecture 6 - 1/14
Sketchingantiderivatives
Basic integrationrules
Problem 19
Compute
∫ 6
−2
2x − 5 dx
by finding areas of regions between the graph of f and the x-axis.
Solution to Problem 19
x-intercept
0 = 2x − 5
x = 52
(area under f above x-axis)
A2 = 12 ·(6− 5
2
)(2 · 6− 5)
= 494
Math 152.02
Lecture 4 - 1/10
Integrals andarea
TheFundamentalTheorem ofCalculus
Interpretingintegrals
Average value ofa function
Lecture 5 - 1/12
Properties ofdefinite integrals
Bounding onintegrals
Antiderivatives
Lecture 6 - 1/14
Sketchingantiderivatives
Basic integrationrules
Solution to Problem 19 (continued)
(area above f under x-axis)
A1 = 12 ·(
52 − (−2)
)(−2 · (−2) + 5)
= 12 ·(
92
)(9)
= 814∫ 6
−2
2x − 5 dx = A2 − A1 = 494 −
814 = −8
Example 20 (Area formulas seldom work)
Evaluate
∫ 3
−5
√25− x2 dx .
No area formula for this portion of a circle
Math 152.02
Lecture 4 - 1/10
Integrals andarea
TheFundamentalTheorem ofCalculus
Interpretingintegrals
Average value ofa function
Lecture 5 - 1/12
Properties ofdefinite integrals
Bounding onintegrals
Antiderivatives
Lecture 6 - 1/14
Sketchingantiderivatives
Basic integrationrules
Theorem 21 (Fundamental Theorem of Calculus I)
If f is continuous on the interval [a, b] and F ′(t) = f (t) then∫ b
a
f (t) dt = F (b)− F (a).
Example 22d
dx
(sin 8x + x4
)= 8 cos 8x + 4x3 so by FTOC∫ 2
1
8 cos 8x + 4x3 dx = (sin(8 · 2) + 24)− (sin(8 · 1) + 14)
= sin 16− sin 8 + 15
Math 152.02
Lecture 4 - 1/10
Integrals andarea
TheFundamentalTheorem ofCalculus
Interpretingintegrals
Average value ofa function
Lecture 5 - 1/12
Properties ofdefinite integrals
Bounding onintegrals
Antiderivatives
Lecture 6 - 1/14
Sketchingantiderivatives
Basic integrationrules
Notation
g(x)∣∣∣bx=a
= g(b)− g(a)
or if x is clearly the variable to plug a and b into can write
g(x)∣∣∣ba
= g(b)− g(a)
With this notation FTOC I (Fundamental Theorem of Calculus I) saysIf f is continuous on the interval [a, b] and F ′(t) = f (t) then∫ b
a
f (t) dt = F (t)∣∣∣bt=a
Math 152.02
Lecture 4 - 1/10
Integrals andarea
TheFundamentalTheorem ofCalculus
Interpretingintegrals
Average value ofa function
Lecture 5 - 1/12
Properties ofdefinite integrals
Bounding onintegrals
Antiderivatives
Lecture 6 - 1/14
Sketchingantiderivatives
Basic integrationrules
FTOC says roughly
Computing areaunder f
≈ Finding a functionwith derivative f
Suprising! Why should area and slopes be related?
Could prove FTOC from definition of integral (Definition 9) anddefinition of the derivative, but we won’t.
Makes some sense visually
Math 152.02
Lecture 4 - 1/10
Integrals andarea
TheFundamentalTheorem ofCalculus
Interpretingintegrals
Average value ofa function
Lecture 5 - 1/12
Properties ofdefinite integrals
Bounding onintegrals
Antiderivatives
Lecture 6 - 1/14
Sketchingantiderivatives
Basic integrationrules
Interpreting integrals
Note on units for integrals
• If x measured in units u1 and
• f (x) measured in units u2 then∫ b
a
f (x) dx
has units u1 · u2
Example 23
• If t is the time measured in hours
• P(t) number of people working in a factory at time t.∫ b
a
P(t) dt
is people · hours worked between time a and b
Math 152.02
Lecture 4 - 1/10
Integrals andarea
TheFundamentalTheorem ofCalculus
Interpretingintegrals
Average value ofa function
Lecture 5 - 1/12
Properties ofdefinite integrals
Bounding onintegrals
Antiderivatives
Lecture 6 - 1/14
Sketchingantiderivatives
Basic integrationrules
Integral of a rate of change
• If F ′(t) is the rate of change of some quantity F (t)
• then FTOC I say that∫ b
a
F ′(t) dt = F (b)− F (a)
This is net change in F from time a to b
Example 24
As in Example 4 from last week
• t time (in h)
• v(t) velocity (rate of change of position) at time t (in km/h)
• then∫ 3
0
v(t) dt
is distance traveled (net change in position) in units(km/h) · h = km
Math 152.02
Lecture 4 - 1/10
Integrals andarea
TheFundamentalTheorem ofCalculus
Interpretingintegrals
Average value ofa function
Lecture 5 - 1/12
Properties ofdefinite integrals
Bounding onintegrals
Antiderivatives
Lecture 6 - 1/14
Sketchingantiderivatives
Basic integrationrules
Example 25
• t time (in years)
• g(t) growth rate (in m/year) of person at age t
• then∫ 16
9
g(t) dt
is change in height (in m) of a person from age 9 to 16.
Math 152.02
Lecture 4 - 1/10
Integrals andarea
TheFundamentalTheorem ofCalculus
Interpretingintegrals
Average value ofa function
Lecture 5 - 1/12
Properties ofdefinite integrals
Bounding onintegrals
Antiderivatives
Lecture 6 - 1/14
Sketchingantiderivatives
Basic integrationrules
Average value of a function
Definition 26 (Average value of a function)
The average value of f on the interval [a, b] is
fave =1
b − a
∫ b
a
f (t) dt.
Where does this formula come from?
Let’s derive it using
Philosophy of Calculus
• Estimate a quantity
• Figure out how to improve estimate
• Take limit
Math 152.02
Lecture 4 - 1/10
Integrals andarea
TheFundamentalTheorem ofCalculus
Interpretingintegrals
Average value ofa function
Lecture 5 - 1/12
Properties ofdefinite integrals
Bounding onintegrals
Antiderivatives
Lecture 6 - 1/14
Sketchingantiderivatives
Basic integrationrules
Average temperature
Let f (t) be the temperature at time t.
Let’s estimate the average temperature between 2am and 10pm.
• (First estimate with 2 intervals)• total time = 10− 2• each interval is 10−2
2•
fave ≈f(2 + 1 · 10−2
2
)+ f
(2 + 2 · 10−2
2
)2
• (Better estimate with 3 intervals)• total time = 10− 2• each interval is 10−2
3•
fave ≈f(2 + 1 · 10−2
3
)+ f
(2 + 2 · 10−2
3
)+ f
(2 + 3 · 10−2
3
)3
Math 152.02
Lecture 4 - 1/10
Integrals andarea
TheFundamentalTheorem ofCalculus
Interpretingintegrals
Average value ofa function
Lecture 5 - 1/12
Properties ofdefinite integrals
Bounding onintegrals
Antiderivatives
Lecture 6 - 1/14
Sketchingantiderivatives
Basic integrationrules
Average temperature (continued)
• (Better estimate with n intervals)• total time = 10− 2• each interval is 10−2
n•
fave ≈f(2 + 1 · 10−2
n
)+ · · ·+ f
(2 + n · 10−2
n
)n
=n∑
i=1
f
(2 + i · 10− 2
n
)· 1n
=1
10− 2
n∑i=1
f
(2 + i · 10− 2
n
)· 10− 2
n
Math 152.02
Lecture 4 - 1/10
Integrals andarea
TheFundamentalTheorem ofCalculus
Interpretingintegrals
Average value ofa function
Lecture 5 - 1/12
Properties ofdefinite integrals
Bounding onintegrals
Antiderivatives
Lecture 6 - 1/14
Sketchingantiderivatives
Basic integrationrules
Average temperature (continued)
• (Take limit to get exact average)
fave = limn→∞
1
10− 2
n∑i=1
f
(2 + i · 10− 2
n
)· 10− 2
n
=1
10− 2lim
n→∞
n∑i=1
f
(2 + i · 10− 2
n
)· 10− 2
n
=1
10− 2
∫ 10
2
f (t) dt
We’ve derived the formula in Definition 26.
Math 152.02
Lecture 4 - 1/10
Integrals andarea
TheFundamentalTheorem ofCalculus
Interpretingintegrals
Average value ofa function
Lecture 5 - 1/12
Properties ofdefinite integrals
Bounding onintegrals
Antiderivatives
Lecture 6 - 1/14
Sketchingantiderivatives
Basic integrationrules
Problem 27
Find the average value of the funtion
h(x) =√
4− x2
on the interval [−2, 2]
Solution to Problem 27
have =1
b − a
∫ b
a
h(x) dx
=1
2− (−2)
∫ 2
−2
√4− x2 dx
=1
4· π · 2
2
2
= π2
Math 152.02
Lecture 4 - 1/10
Integrals andarea
TheFundamentalTheorem ofCalculus
Interpretingintegrals
Average value ofa function
Lecture 5 - 1/12
Properties ofdefinite integrals
Bounding onintegrals
Antiderivatives
Lecture 6 - 1/14
Sketchingantiderivatives
Basic integrationrules
Problem 28
• t is time measured in days since Jan. 1, 2003
• R(t) is the distance from the earth to the sun at time t
What does1
365− 0
∫ 365
0
R(t) dt
represent?
Solution to Problem 28
Average distance from earth to the sun in 2003
Math 152.02
Lecture 4 - 1/10
Integrals andarea
TheFundamentalTheorem ofCalculus
Interpretingintegrals
Average value ofa function
Lecture 5 - 1/12
Properties ofdefinite integrals
Bounding onintegrals
Antiderivatives
Lecture 6 - 1/14
Sketchingantiderivatives
Basic integrationrules
Properties of definite integrals
Theorem 29
If f is integrable then∫ b
a
f (x) dx = −∫ a
b
f (x) dx
Theorem 30
If f is integrable then∫ b
a
f (x) dx +
∫ c
b
f (x) dx =
∫ c
a
f (x) dx
Math 152.02
Lecture 4 - 1/10
Integrals andarea
TheFundamentalTheorem ofCalculus
Interpretingintegrals
Average value ofa function
Lecture 5 - 1/12
Properties ofdefinite integrals
Bounding onintegrals
Antiderivatives
Lecture 6 - 1/14
Sketchingantiderivatives
Basic integrationrules
Theorem 31
If f is integrable then∫ b
a
cf (x) dx = c
∫ b
a
f (x) dx
Proof. ∫ b
a
cf (x) dx = limn→∞
n∑i=1
cf (a + i∆x)∆x
= limn→∞
cn∑
i=1
f (a + i∆x)∆x
= c limn→∞
n∑i=1
f (a + i∆x)∆x
= c
∫ b
a
f (x) dx
Math 152.02
Lecture 4 - 1/10
Integrals andarea
TheFundamentalTheorem ofCalculus
Interpretingintegrals
Average value ofa function
Lecture 5 - 1/12
Properties ofdefinite integrals
Bounding onintegrals
Antiderivatives
Lecture 6 - 1/14
Sketchingantiderivatives
Basic integrationrules
Theorem 32
If f and g are integrable then∫ b
a
f (x) + g(x) dx =
∫ b
a
f (x) dx +
∫ b
a
g(x) dx
Proof.∫ b
a
f (x) + g(x) dx = limn→∞
n∑i=1
(f (a + i∆x) + g(a + i∆x)) ∆x
= limn→∞
n∑i=1
(f (a + i∆x)∆x + g(a + i∆x)∆x)
Math 152.02
Lecture 4 - 1/10
Integrals andarea
TheFundamentalTheorem ofCalculus
Interpretingintegrals
Average value ofa function
Lecture 5 - 1/12
Properties ofdefinite integrals
Bounding onintegrals
Antiderivatives
Lecture 6 - 1/14
Sketchingantiderivatives
Basic integrationrules
Proof of Theorem 32 (continued).
= limn→∞
n∑i=1
(f (a + i∆x)∆x + g(a + i∆x)∆x)
= limn→∞
(n∑
i=1
f (a + i∆x)∆x +n∑
i=1
g(a + i∆x)∆x
)
=
(lim
n→∞
n∑i=1
f (a + i∆x)∆x
)+
(lim
n→∞
n∑i=1
g(a + i∆x)∆x
)
=
∫ b
a
f (x) dx +
∫ b
a
g(x) dx
Math 152.02
Lecture 4 - 1/10
Integrals andarea
TheFundamentalTheorem ofCalculus
Interpretingintegrals
Average value ofa function
Lecture 5 - 1/12
Properties ofdefinite integrals
Bounding onintegrals
Antiderivatives
Lecture 6 - 1/14
Sketchingantiderivatives
Basic integrationrules
Problem 33
Compute∫ 16
7t(v)− 7u(v) dv given
•∫ 16
7u(v) dv = −2
•∫ 16
7t(v) dv = 12
Solution to Problem 33∫ 16
7
t(v)− 7u(v) dv =
∫ 16
7
t(v) dv −∫ 16
7
7u(v) dv
=
∫ 16
7
t(v) dv − 7
∫ 16
7
u(v) dv
= 12− 7(−2)
= 26
Math 152.02
Lecture 4 - 1/10
Integrals andarea
TheFundamentalTheorem ofCalculus
Interpretingintegrals
Average value ofa function
Lecture 5 - 1/12
Properties ofdefinite integrals
Bounding onintegrals
Antiderivatives
Lecture 6 - 1/14
Sketchingantiderivatives
Basic integrationrules
Warning
∫ b
a
f (x) · g(x) dx 6=∫ b
a
f (x) dx ·∫ b
a
g(x) dx
∫ b
a
f (x)
g(x)dx 6=
∫ b
af (x) dx∫ b
ag(x) dx
Example 34
f (x) =
{1, 0 ≤ x ≤ 10, 1 < x ≤ 2
and g(x) =
{0, 0 ≤ x ≤ 11, 1 < x ≤ 2
∫ 2
0
f (x)g(x) dx =
∫ 2
0
0 dx = 0
but
∫ 2
0
f (x) dx ·∫ 2
0
g(x) dx = 1 · 1 = 1
Math 152.02
Lecture 4 - 1/10
Integrals andarea
TheFundamentalTheorem ofCalculus
Interpretingintegrals
Average value ofa function
Lecture 5 - 1/12
Properties ofdefinite integrals
Bounding onintegrals
Antiderivatives
Lecture 6 - 1/14
Sketchingantiderivatives
Basic integrationrules
Definition 35
• f is even if f (−x) = f (x)
• f is odd if f (−x) = −f (x)
Note: Most functions are neither. One function is both.
Theorem 36
If f is even then∫ a
0
f (x) dx =
∫ 0
−af (x) dx = 1
2
∫ a
−af (x) dx
If f is odd then ∫ a
0
f (x) dx = −∫ 0
−af (x) dx
and ∫ a
−af (x) dx = 0
Math 152.02
Lecture 4 - 1/10
Integrals andarea
TheFundamentalTheorem ofCalculus
Interpretingintegrals
Average value ofa function
Lecture 5 - 1/12
Properties ofdefinite integrals
Bounding onintegrals
Antiderivatives
Lecture 6 - 1/14
Sketchingantiderivatives
Basic integrationrules
Problem 37
Suppose
•∫ 4
2Q(s) ds = 3
•∫ 20
4Q(s) ds = −9
Evaluate ∫ 20
2
Q(s) ds
Solution to Problem 37
∫ 20
2
Q(s) ds =
∫ 4
2
Q(s) ds +
∫ 20
4
Q(s) ds
= 3 + (−9)
= −6
Math 152.02
Lecture 4 - 1/10
Integrals andarea
TheFundamentalTheorem ofCalculus
Interpretingintegrals
Average value ofa function
Lecture 5 - 1/12
Properties ofdefinite integrals
Bounding onintegrals
Antiderivatives
Lecture 6 - 1/14
Sketchingantiderivatives
Basic integrationrules
Problem 38
Compute∫ 8
2y(u) du given
• y is an odd function
•∫ 8
−2y(u) du = 17
Solution to Problem 38∫ 8
2
y(u) du =
∫ 8
−2
y(u) du −∫ 2
−2
y(u) du
= 17− 0
= 17
Math 152.02
Lecture 4 - 1/10
Integrals andarea
TheFundamentalTheorem ofCalculus
Interpretingintegrals
Average value ofa function
Lecture 5 - 1/12
Properties ofdefinite integrals
Bounding onintegrals
Antiderivatives
Lecture 6 - 1/14
Sketchingantiderivatives
Basic integrationrules
Problem 39
Compute∫ 7
3h(r) dr given
• h is an even function
•∫ 7
0h(r) dr = 4
•∫ 7
−3h(r) dr = −10
Solution to Problem 39∫ 7
3
h(r) dr =
∫ 7
−3
h(r) dr −∫ 3
−3
h(r) dr
=
∫ 7
−3
h(r) dr − 2
∫ 0
−3
h(r) dr
=
∫ 7
−3
h(r) dr − 2
(∫ 7
−3
h(r) dr −∫ 7
0
h(r) dr
)= −
∫ 7
−3
h(r) dr + 2
∫ 7
0
h(r) dr
= −(−10) + 2(4) = 18
Math 152.02
Lecture 4 - 1/10
Integrals andarea
TheFundamentalTheorem ofCalculus
Interpretingintegrals
Average value ofa function
Lecture 5 - 1/12
Properties ofdefinite integrals
Bounding onintegrals
Antiderivatives
Lecture 6 - 1/14
Sketchingantiderivatives
Basic integrationrules
Bounding on integrals
Theorem 40
If m ≤ f (x) ≤ M for x ∈ [a, b] then
m(b − a) ≤∫ b
a
f (x) dx ≤ M(b − a)
Theorem 41
If f (x) ≤ g(x) for x ∈ [a, b] then∫ b
a
f (x) dx ≤∫ b
a
g(x) dx
Math 152.02
Lecture 4 - 1/10
Integrals andarea
TheFundamentalTheorem ofCalculus
Interpretingintegrals
Average value ofa function
Lecture 5 - 1/12
Properties ofdefinite integrals
Bounding onintegrals
Antiderivatives
Lecture 6 - 1/14
Sketchingantiderivatives
Basic integrationrules
Problem 42
Show that
−199 ≤∫ 200
1
cos(x) · sin(x3) dx ≤ 199
Solution to Problem 42
For all x ∈ [1, 200]
| cos(x) · sin(x3)| = | cos(x)| · | sin(x3)| ≤ 1
so−1 ≤ cos(x) · sin(x3) ≤ 1
By Theorem 40
−1 · (200− 1) ≤∫ 200
1
cos(x) · sin(x3) dx ≤ 1 · (200− 1)
−199 ≤∫ 200
1
cos(x) · sin(x3) dx ≤ 199
Math 152.02
Lecture 4 - 1/10
Integrals andarea
TheFundamentalTheorem ofCalculus
Interpretingintegrals
Average value ofa function
Lecture 5 - 1/12
Properties ofdefinite integrals
Bounding onintegrals
Antiderivatives
Lecture 6 - 1/14
Sketchingantiderivatives
Basic integrationrules
Problem 43
For each pairs of integrals decide which is the larger
1∫ π
4
0cos(x) dx and
∫ π4
0sin(x) dx
2∫ π
2π4
cos(x) dx and∫ π
2π4
sin(x) dx
Solution to Problem 43
1 For all x ∈ [0, π4 ]cos(x) ≥ sin(x)
so ∫ π4
0
cos(x) dx ≥∫ π
4
0
sin(x) dx
2 For all x ∈ [π4 ,π2 ]
cos(x) ≤ sin(x)
so ∫ π2
π4
cos(x) dx ≤∫ π
2
π4
sin(x) dx
Math 152.02
Lecture 4 - 1/10
Integrals andarea
TheFundamentalTheorem ofCalculus
Interpretingintegrals
Average value ofa function
Lecture 5 - 1/12
Properties ofdefinite integrals
Bounding onintegrals
Antiderivatives
Lecture 6 - 1/14
Sketchingantiderivatives
Basic integrationrules
Antiderivatives
Definition 44 (An antiderivative)
F (x) is an antiderivative of f (x) if F ′(x) = f (x).
Example 45
ddx
(x2 cos(4x + 3)
)= 2x cos(4x + 3) + 4x2 sin(4x + 3)
sox2 cos(4x + 3)
is an antiderivative of
2x cos(4x + 3) + 4x2 sin(4x + 3)
x2 cos(4x + 3) + 30 is another antiderivative.
Math 152.02
Lecture 4 - 1/10
Integrals andarea
TheFundamentalTheorem ofCalculus
Interpretingintegrals
Average value ofa function
Lecture 5 - 1/12
Properties ofdefinite integrals
Bounding onintegrals
Antiderivatives
Lecture 6 - 1/14
Sketchingantiderivatives
Basic integrationrules
Motivation
Why do we care about finding antiderivatives?
FTOC I says that computing ∫ b
a
f (t) dt
is easy if we have an antiderivative F of f .
Example 46
From Example 45 above∫ 7
1
2x cos(4x + 3) + 4x2 sin(4x + 3) dx
= x2 cos(4x + 3) + 30∣∣∣71
= (72 cos(4 · 7 + 3) + 30)− (12 cos(4 · 1 + 3) + 30)
= 49 cos(31)− cos(7)
Math 152.02
Lecture 4 - 1/10
Integrals andarea
TheFundamentalTheorem ofCalculus
Interpretingintegrals
Average value ofa function
Lecture 5 - 1/12
Properties ofdefinite integrals
Bounding onintegrals
Antiderivatives
Lecture 6 - 1/14
Sketchingantiderivatives
Basic integrationrules
Notice in Example 45 we could have added any constant tox2 cos(4x + 3) and we would have had another antiderivative of2x cos(4x + 3) + 4x2 sin(4x + 3)
We usually add an unspecified constant to remind us that there aremany antiderivatives.
Definition 47 (The antiderivative)
The antiderivative of f (x) is the set of all antiderivatives of f (x).
Theorem 48
If f is continuous and F ′(x) = f (x) then every antiderivative of f isof the form
F (x) + C
for some constant C .
Math 152.02
Lecture 4 - 1/10
Integrals andarea
TheFundamentalTheorem ofCalculus
Interpretingintegrals
Average value ofa function
Lecture 5 - 1/12
Properties ofdefinite integrals
Bounding onintegrals
Antiderivatives
Lecture 6 - 1/14
Sketchingantiderivatives
Basic integrationrules
What if f is not continuous?
The antiderivative of a noncontinuous function
Let
F (x) =
{ln x + 14, x > 0ln(−x) + 8, x < 0
Then
F ′(x) =
{1x , x > 0−1−x , x < 0
=
{1x , x > 01x , x < 0
=1
x
So F (x) is an antiderivative of 1x .
Any choice of constants (14 and 8 weren’t special) gives same result.
Thus the antiderivative of 1x is
F (x) =
{ln x + C1, x > 0ln(−x) + C2, x < 0
=
{ln |x |+ C1, x > 0ln |x |+ C2, x < 0
Math 152.02
Lecture 4 - 1/10
Integrals andarea
TheFundamentalTheorem ofCalculus
Interpretingintegrals
Average value ofa function
Lecture 5 - 1/12
Properties ofdefinite integrals
Bounding onintegrals
Antiderivatives
Lecture 6 - 1/14
Sketchingantiderivatives
Basic integrationrules
On the other hand
• Main reason we care about antiderivatives is the FTOC.
• FTOC only applies if f is integrable on [a, b]
• 1x is not integrable on intervals containing 0
• so in applications we only use one of the two constants at a time
Example 49∫ −2
−3
1
xdx = ln |x |+C2
∣∣∣−2
−3= (ln |−2|+C2)−(ln |−3|+C2) = ln 2−ln 3
Example 50∫ 7
−4
1
xdx cannot be evaluated using FTOC
Notational warning
By convention we say that F (x) + C is the antiderivative of f (x)whenever F ′(x) = f (x) even when this is technically incorrect.
Math 152.02
Lecture 4 - 1/10
Integrals andarea
TheFundamentalTheorem ofCalculus
Interpretingintegrals
Average value ofa function
Lecture 5 - 1/12
Properties ofdefinite integrals
Bounding onintegrals
Antiderivatives
Lecture 6 - 1/14
Sketchingantiderivatives
Basic integrationrules
Notation ∫f (x) dx = F (x) + C
means F (x) + C is the antiderivative of f (x)
Terminology
Since FTOC links antidifferentiation and integration we also callantiderivatives (indefinite) integrals.
The following statements all mean the same thing:
• f (x) = ddx F (x)
•∫
f (x) dx = F (x) + C
• f (x) is the derivative of F (x)
• F (x) + C is the antiderivative of f (x)
• F (x) + C is the indefinite integral of f (x)
• F (x) + C is the integral of f (x)
Math 152.02
Lecture 4 - 1/10
Integrals andarea
TheFundamentalTheorem ofCalculus
Interpretingintegrals
Average value ofa function
Lecture 5 - 1/12
Properties ofdefinite integrals
Bounding onintegrals
Antiderivatives
Lecture 6 - 1/14
Sketchingantiderivatives
Basic integrationrules
Problem 51
Check the following integrals
1∫
(6x + 3ex) · cos(3x2 + 3ex) dx = sin(3x2 + 3ex) + C
2∫
sec x dx = ln | sec x + tan x |+ C
Solution to Problem 51
1d
dx sin(3x2 + 3ex) = (6x + 3ex) · cos(3x2 + 3ex) X
2 ddx ln | sec x + tan x | =
sec x tan x + sec2 x
sec x + tan x
= (sec x) · tan x + sec x
sec x + tan x
= sec x XNote: Similar to 1
x antiderivative of sec x should have a different
C for each interval [ (2n−1)π2 , (2n+1)π
2 ] but nobody does this.
Math 152.02
Lecture 4 - 1/10
Integrals andarea
TheFundamentalTheorem ofCalculus
Interpretingintegrals
Average value ofa function
Lecture 5 - 1/12
Properties ofdefinite integrals
Bounding onintegrals
Antiderivatives
Lecture 6 - 1/14
Sketchingantiderivatives
Basic integrationrules
(See backboard)
Math 152.02
Lecture 4 - 1/10
Integrals andarea
TheFundamentalTheorem ofCalculus
Interpretingintegrals
Average value ofa function
Lecture 5 - 1/12
Properties ofdefinite integrals
Bounding onintegrals
Antiderivatives
Lecture 6 - 1/14
Sketchingantiderivatives
Basic integrationrules
Basic integration rules
Each rule for differentiation gives us a rule for integration
Fromc d
dx F (x) = ddx
(cF (x)
)we get
Theorem 52 (Constant rule for integration)
∫cf (x) dx = c
∫f (x) dx
Math 152.02
Lecture 4 - 1/10
Integrals andarea
TheFundamentalTheorem ofCalculus
Interpretingintegrals
Average value ofa function
Lecture 5 - 1/12
Properties ofdefinite integrals
Bounding onintegrals
Antiderivatives
Lecture 6 - 1/14
Sketchingantiderivatives
Basic integrationrules
Proof of Theorem 52.
Suppose ddx F (x) = f (x).
We have the derivative rule
c ddx F (x) = d
dx
(cF (x)
)Reinterpreting this rule as an antiderivative gives∫
c ddx F (x) dx = cF (x) + C .
Thus we may conclude∫cf (x) dx =
∫c d
dx F (x) dx
= cF (x) + C
= c(F (x) + C2)
= c
∫f (x) dx .
Math 152.02
Lecture 4 - 1/10
Integrals andarea
TheFundamentalTheorem ofCalculus
Interpretingintegrals
Average value ofa function
Lecture 5 - 1/12
Properties ofdefinite integrals
Bounding onintegrals
Antiderivatives
Lecture 6 - 1/14
Sketchingantiderivatives
Basic integrationrules
From ddx F (x) + d
dx G (x) = ddx
(F (x) + G (x)
)we get
Theorem 53 (Sum rule for integration)
∫f (x) + g(x) dx =
∫f (x) dx +
∫g(x) dx
Math 152.02
Lecture 4 - 1/10
Integrals andarea
TheFundamentalTheorem ofCalculus
Interpretingintegrals
Average value ofa function
Lecture 5 - 1/12
Properties ofdefinite integrals
Bounding onintegrals
Antiderivatives
Lecture 6 - 1/14
Sketchingantiderivatives
Basic integrationrules
Proof of Theorem 53.
Suppose ddx F (x) = f (x) and d
dx G (x) = g(x).
We have the derivative rule
ddx F (x) + d
dx G (x) = ddx
(F (x) + G (x)
)Reinterpreting this rule as an antiderivative gives∫
ddx F (x) + d
dx G (x) dx = F (x) + G (x) + C .
Thus we may conclude∫f (x) + g(x) dx =
∫d
dx F (x) + ddx G (x) dx
= F (x) + G (x) + C
=
∫f (x) dx +
∫g(x) dx
Note: We drop constants when we have integrals on both sides of anequation.
Math 152.02
Lecture 4 - 1/10
Integrals andarea
TheFundamentalTheorem ofCalculus
Interpretingintegrals
Average value ofa function
Lecture 5 - 1/12
Properties ofdefinite integrals
Bounding onintegrals
Antiderivatives
Lecture 6 - 1/14
Sketchingantiderivatives
Basic integrationrules
Basic Integrals
Each basic derivative gives us a basic integral
Table 1: Basic integrals to memorize
differentiation integrationrule rule
ddx x r+1 = (r + 1)x r
∫x r dx = 1
r+1 x r+1 + C if r 6= −1
ddx ln |x | = 1
x
∫1x dx = ln |x |+ C
ddx cos x = − sin x
∫sin x dx = − cos x + C
ddx sin x = cos x
∫cos x dx = sin x + C
ddx ex = ex
∫ex dx = ex + C
ddx arctan x = 1
1+x2
∫1
1+x2 dx = arctan x + C
ddx arcsin x = 1√
1−x2
∫1√
1−x2dx = arcsin x + C
Math 152.02
Lecture 4 - 1/10
Integrals andarea
TheFundamentalTheorem ofCalculus
Interpretingintegrals
Average value ofa function
Lecture 5 - 1/12
Properties ofdefinite integrals
Bounding onintegrals
Antiderivatives
Lecture 6 - 1/14
Sketchingantiderivatives
Basic integrationrules
More basic integrals
You also know a few more derivative rules
Table 2: More basic integrals to memorize
differentiation integrationrule rule
ddx tan x = sec2 x
∫sec2 x dx = tan x + C
ddx cot x = − csc2 x
∫csc2 x dx = − cot x + C
ddx sec x = sec x tan x
∫sec x tan x dx = sec x + C
ddx csc x = − csc x cot x
∫csc x cot x dx = − csc x + C
ddx ax = (ln a)ax
∫ax dx = ax
ln a + C
Math 152.02
Lecture 4 - 1/10
Integrals andarea
TheFundamentalTheorem ofCalculus
Interpretingintegrals
Average value ofa function
Lecture 5 - 1/12
Properties ofdefinite integrals
Bounding onintegrals
Antiderivatives
Lecture 6 - 1/14
Sketchingantiderivatives
Basic integrationrules
Techniques of integration
Advanced derivative rules give us techniques of integration
differentiation technique ofrule integration
chain rule u-substitution (§7.1)
product rule integration by parts (§7.2)
We will return to these integration techniques later.
Math 152.02
Lecture 4 - 1/10
Integrals andarea
TheFundamentalTheorem ofCalculus
Interpretingintegrals
Average value ofa function
Lecture 5 - 1/12
Properties ofdefinite integrals
Bounding onintegrals
Antiderivatives
Lecture 6 - 1/14
Sketchingantiderivatives
Basic integrationrules
Problem 54
Find a formula for∫
10ex + 7 sin x dx
Solution to Problem 54∫10ex + 7 sin x dx =
∫10ex dx +
∫7 sin x dx (Sum rule)
= 10
∫ex dx + 7
∫sin x dx (Constant rule)
= 10ex − 7 cos x + C (Table 1)
Check your answer!
ddx (10ex − 7 cos x) = 10ex + 7 sin x X
Math 152.02
Lecture 4 - 1/10
Integrals andarea
TheFundamentalTheorem ofCalculus
Interpretingintegrals
Average value ofa function
Lecture 5 - 1/12
Properties ofdefinite integrals
Bounding onintegrals
Antiderivatives
Lecture 6 - 1/14
Sketchingantiderivatives
Basic integrationrules
Problem 55
Find a formula for∫
18√1−t2− 8t22 dt
Solution to Problem 55∫18√
1− t2− 8t22 dt =
∫18 · 1√
1− t2dt −
∫8 · t22 dt (Sum rule)
= 18
∫1√
1− t2dt − 8
∫t22 dt (Const. rule)
= 18 arcsin t − 8 · t23
23 + C (Table 1)
Check your answer!
ddt
(18 arcsin t − 8
23 · t23)
=18√
1− t2− 8
23 ·23t22 =18√
1− t2−8t22 X
Math 152.02
Lecture 4 - 1/10
Integrals andarea
TheFundamentalTheorem ofCalculus
Interpretingintegrals
Average value ofa function
Lecture 5 - 1/12
Properties ofdefinite integrals
Bounding onintegrals
Antiderivatives
Lecture 6 - 1/14
Sketchingantiderivatives
Basic integrationrules
Problem 56
Find a formula for∫ arcsin(3−π2)√
2+ sec2 u du
Solution to Problem 56
∫arcsin(3− π2)√
2+ cos u du
=
∫arcsin(3− π2)√
2du +
∫sec2 u du (Sum rule)
= arcsin(3−π2)√2
· u + tan u + C (Tables 1 and 2)
Check your answer!
ddt
(arcsin(3− π2)√
2· u + tan u
)=
arcsin(3− π2)√2
+ sec2 u X
Math 152.02
Lecture 4 - 1/10
Integrals andarea
TheFundamentalTheorem ofCalculus
Interpretingintegrals
Average value ofa function
Lecture 5 - 1/12
Properties ofdefinite integrals
Bounding onintegrals
Antiderivatives
Lecture 6 - 1/14
Sketchingantiderivatives
Basic integrationrules
Problem 57
Compute∫ 2
1
3√y(1−6 6
√y)2
3√y dy
Solution to Problem 57∫ 2
1
3√
y(1− 6 6√
y)2
3√
ydy =
∫ 2
1
3y12 (1− 12y
16 + 36y
26 )
y13
dy
=
∫ 2
1
3y12 − 36y
46 + 108y
56 )
y13
dy
=
∫ 2
1
3y16 − 36y
26 + 108y
36 dy
=
∫ 2
1
3y16 − 36y
13 + 108y
12 dy
= 3 · 67 y
76 − 36 · 3
4 y43 + 108 · 2
3 y32
∣∣∣21
= 187 y
76 − 27y
43 + 72y
32
∣∣∣21
=(
187 · 2
76 − 27 · 2 4
3 + 72 · 2 32
)−(
187 + 45
)
Math 152.02
Lecture 4 - 1/10
Integrals andarea
TheFundamentalTheorem ofCalculus
Interpretingintegrals
Average value ofa function
Lecture 5 - 1/12
Properties ofdefinite integrals
Bounding onintegrals
Antiderivatives
Lecture 6 - 1/14
Sketchingantiderivatives
Basic integrationrules
Problem 58
Find an antiderivative G (x) of g(x) = sin x + 7 satisfyingG (π) = −20.
Solution to Problem 58
G (x) =
∫sin x + 7 dx
= − cos x + 7x + C
Use fact that G (π) = −20 to solve for C .
−20 = G (π) = − cosπ + C
SoC = −20 + cosπ = −20 + (−1) = −21
G (x) = − cos x + 7x − 21
Math 152.02
Lecture 4 - 1/10
Integrals andarea
TheFundamentalTheorem ofCalculus
Interpretingintegrals
Average value ofa function
Lecture 5 - 1/12
Properties ofdefinite integrals
Bounding onintegrals
Antiderivatives
Lecture 6 - 1/14
Sketchingantiderivatives
Basic integrationrules
Problem 59
The average value of h(x) = x3 − 3x2 on [−a, a] is 8 solve for a.
Solution to Problem 59
have =1
a− (−a)
∫ a
−ax3 − 3x2 dx
=1
2a· ( 1
4 x4 − x3)∣∣∣a−a
=1
2a·(
[ 14 (−a)4 − (−a)3]− ( 1
4 a4 − a3))
=1
2a·(
14 a4 + a3 − 1
4 a4 + a3)
=1
2a·(2a3)
= a2
Use fact that have = 8 to solve for a.
8 = have = a2 so a = ±√
8