110A Chapter 5 Integrals 12-14-17 - WordPress.com
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Chapter 5 – Integrals 5.1 Areas and Distances We start with a problem – how can we calculate the area “under” a given function – ie, the area between the function and the x-axis? If the curve happens to be something easy – like a horizontal line - this isn’t too hard, but if it’s a curve, the problem becomes more difficult. Let’s start by drawing a rectangle that’s f(a) tall by (b – a) wide as an estimate: Area = f(a) (b – a) How can we get a better estimate? Let’s try 2 rectangles: Area = f(a) ∆x 1 + f(c) ∆x 2 How could we get an even better estimate? HT µ § flat ← bit T.x.tk
Transcript of 110A Chapter 5 Integrals 12-14-17 - WordPress.com
Chapter 5 – Integrals 5.1 Areas and Distances We start with a problem – how can we calculate the area “under” a given function – ie, the area between the function and the x-axis? If the curve happens to be something easy – like a horizontal line - this isn’t too hard, but if it’s a curve, the problem becomes more difficult.
Let’s start by drawing a rectangle that’s f(a) tall by (b – a) wide as an estimate:
Area = f(a) (b – a) How can we get a better estimate? Let’s try 2 rectangles:
Area = f(a) ∆x1 + f(c) ∆x2 How could we get an even better estimate?
HT
µ § flat
←
bitT.x.tk
Ex: Estimate the area under the curve f(x) = x2 between x = 0 and x = 1 by constructing 8 rectangles (the book calls them “strips”) of equal width. This seems like a relatively simple situation, until you realize that you get different answers depending on whether you draw the rectangles from the left or from the right:
Using left endpoints, we get: f(0) !
" + f !"
!" + f #
"!" + f $
"!" + f %
"!" + f &
"!" + f '
"!" + f (
"!"
We can see this will be a lower bound, because all the rectangles fit completely under the curve. Using right endpoints, we get: f !"
!" + f #
"!" + f $
"!" + f %
"!" + f &
"!" + f '
"!" + f (
"!" + f 1 !
" We can see this will be an upper bound, because all the rectangles extend to above the curve. By computing both, we get: 0.2734375 < A < 0.3984375 Hmmm…if only we had a technique for looking at a progression of smaller and smaller rectangles….
%.
n= 8 ~ n't
I ax I ax
Left Right
Definition: The Area A of the region S that lies under the graph of the continuous function f is the limit of the sum of the areas of approximating rectangles (using right endpoints): A = lim
-→/0- = lim-→/ 3 4! ∆4 + 3 4# ∆4 + ⋯+ 3(4-)∆4
1) It can be proven that this limit always exists. 2) It can also be shown that you get the same limit value A from using left endpoints. 3) In fact, you can choose any x-value in each interval to calculate the height. The values that are chosen - x1
*, x2*, …, xn
* - are called the sample points. Using sigma notation we get: A = lim
-→/3(4:)∆4-
:;! (Riemann sum) The Distance Problem My odometer is broken, but I’d still like to calculate how far I’ve driven using my speedometer and a stopwatch (held by my passenger, for safety.) I know D = R*T and I have the following data:
Time (s) 0 5 10 15
Speed (ft/s) 0 20 25 10 How can I estimate the distance I’ve travelled?
" = # " " " *
F)× ,* -
ax
=hig•§fCxDa×
If .
Using Left Endpoints: 0 * 5 + 20 * 5 + 25 * 5 = 225 ft Using Right Endpoints: 20 * 5 + 25 * 5 + 10 * 5 = 275 ft 1) How can we improve our estimate? 2) Do we know whether the true distance lies between the two estimates?
=More data points
smaller ax
## non"
5.2 The Definite Integral The limit of the sum from 5.1: lim-→/
3 4! ∆4 + 3 4# ∆4 + ⋯+ 3(4-)∆4 = lim-→/
3(4:)∆4-:;!
where ∆4 = <=>- , can be written as:
3 4 ?4<
>
and is called the definite integral of f from a to b, as long as the limit exists and is the same for all choices of xi. If the limit exists, then f is said to be integrable over [a,b]
Properties of the Integral 1) @?4<
> = c(b – a) where c is any constant 2) 3 4 + A 4 ?4 =<
> 3 4 ?4<> + A 4 ?4<
> 3) @3 4 ?4<
> = c 3 4 ?4<> where c is any constant
4) [3 4 − A 4 ]?4<
> = 3 4 ?4<> - A 4 ?4<
> 5) 3 4 ?4E
> + 3 4 ?4<E = 3 4 ?4<
> 6) If f(x) > 0 for a < x < b, then 3 4 ?4<
> > 0 7) If f(x) > g(x) for a < x < b, then 3 4 ?4<
> > A 4 ?4<>
8) If m < f(x) < M for a < x < b, then m (b – a) < 3 4 ?4<
> ≤ G(H − I)
fsumas i. 1,33,
...n
Ex: (#33) The graph of f is shown. Evaluate each integral by interpreting it in terms of areas.
a) 3 4 ?4#
J d) 3 4 ?4K
J
÷ 5
inP8f÷,I¥ pain n€ay
.
=3+ It 'z= 4
= (10) + ( -8 ) = 2
5.3 The Fundamental Theorem of Calculus (Cue exciting music)
The Fundamental Theorem of Calculus, Part I If f is continuous on [a,b], then the function g defined by g(x) = 3 L ?LM
> a < x < b is continuous on [a,b] and differentiable on (a,b), and g’(x) = f(x).
Ex: Find NNM sec L?LMR
! (Hint: You’ll need the Chain Rule)
The Fundamental Theorem of Calculus, Part 2 If f is continuous on [a, b], then 3 4 ?4<
> = F(b) – F(a) where F is any antiderivative of f. That is, a function s. t. F’ = f.
Why can it be *any* antiderivative of f?
( Ex4)
£, f.
*sect dt = see (x 4) 4×3
So, differentiation and integration are inverse processes. Ex: Evaluate these integrals 1) 1 + 4 + 34# ?4!
J 2) 4 − L L?L%
J Note: Be careful when rushing in to evaluate integrals. Look at this example:
14# ?4
$
=!
It seems like it would be easy to find an antiderivative and evaluate it, but first – is this function continuous over [-1, 3]?
=x+Iz+§II=( it '÷+P ) . (0+0*03)
= ( z 's ) -0=2 'EorE
[email protected],ij⇐⇐f¥¥I¥I¥l¥¥f'
= Z+l=3ztE=E =§ . 8- ¥32= 6¥ . 651=3,25-0*2
= 12815
=[ xtdx
ifeng.to
5.4 Indefinite Integrals Definite vs. Indefinite Integrals: 24?4$
J is a definite integral. It’s a string of symbols that represent a number: 24?4$
J = x2 ]03 = 32 – 02 = 9 24?4is an indefinite integral. It’s a string of symbols that represent a
function, or family of functions: 24?4= x2 + C
Table of Indefinite Integrals I 3 4 ?4 =a 3(4) ?4
∫ [f(x) + g(x)] dx = ∫ 3 4 ?4 +∫ A 4 ?4 ∫ W?4 = kx + C ∫ xndx = M
XYZ
-[! + C (n ≠ -1) ∫ ex dx = ex + C ∫ !
M dx = ln |x| + C ∫ cos x dx = sinx + C ∫ sin x dx = - cos x + C ∫ sec2 x dx = tan x + C ∫ sec x tan x dx = sec x + C
⇐t€ta¥¥o⇒
Net Change Theorem The integral of a rate of change is the net change
\] 4 ?4 = \ H − \(I)<
>
gab (Position
) #Speed
5.5 The Substitution Rule Ex: a) Find NNM (4
# + 34)$ b) Find 4# + 34 # 64 + 9 ?4
The Substitution Rule If u = g(x) is a differentiable function whose range is an interval I and f is continuous on I, then 3 A 4 A] 4 ?4 = 3 ` ?`
Ex: Evaluate 24 + 1 ?4 using the Substitution Rule
⇒x2t3x5 (2×+3) ( x2t3× )
'
t.CI#6et9) Sf )£,
v. (12+3×56×+9)
= (#D3tCWhat if we didn't just ]
do porta ?
f(×'+3×56×+9 )dx
Let U=×43Xdu=( 2×+31 ?
f(u)23du= 03.3
'
Itc=i(x43D'tC
Let u= 2×+1 ( ¥×=2 )du=Zdx
d÷=dxfo¥=±Su÷du=E¥±+C=⇒u¥c=±u÷+c=÷(zx+DE+C
Ex : ) ( ×2+5x)2
( zxt 5) dx
Letu=×2t5Xdu= (2×+5) dx
= f u2du
= Is + C
= ( ×45x)33-
+ C
ex : SEIF = St ¥Let u=x2t4 = fjtdydu= Zxdx
d÷=xdx -
.
Ex: Evaluate 1 + ##!
J #&)# using the Substitution Rule Integrals of Symmetric Functions – If f is even [ f(-x) = f(x) ], then * # )#>
=> = 2 * # )#>J
If f is odd [ f(-x) = -f(x) ], then * # )#>
=> = 0
Let u=1+×2 du= Zxdx
U - 1=×2 df=×d×
§Fttx5dx=§FxdxYxdx
=fhF(u-D'¥={F0?zu+D¥-
fahfcitu + D ¥
=Sub bEzu±+u⇒ E
= he . ¥÷¥t±I= ⇐E . 4y±+ }u÷ ] . 's ] != ¥ . 3¥ + ¥3 !
= at¥EyEIEC¥÷]j= ¥. 2¥ +3¥ ( t . } ⇒
= 8¥ - 8¥ + 2¥ - t+¥3→
= 8¥ - 8¥ +2¥ - ttf - st
=EC÷ . Its ) - E.
t.no#s5=rC*iEt*s.oa=r(FEt@
