Section 17.4Integration
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A physicist who knows the velocity of a particle might wish to know its position at a given time.
A biologist who knows the rate at which a bacteria population is increasing might want to deduce what the size of the population will be at some future time.
Introduction
In each case, the problem is to find a function F whose derivative is a known function f.
If such a function F exists, it is called an antiderivative of f.
Antiderivatives
Definition
A function F is called an antiderivative of f on
an interval I if F’(x) = f (x) for all x in I.
Find the integral. (Find the antiderivative.)
= ? dxx n
€
1n +1 x n +1 + C
If F is an antiderivative of f on an interval I, then the most general antiderivative of f on I is
F(x) + C
where C is an arbitrary constant.
Theorem
Antiderivatives
Going back to the function f (x) = x2, we see that the general antiderivative of f is ⅓ x3 + C.
Notation for Antiderivatives
The symbol is traditionally used to represent the most general an antiderivative of f on an open interval and is called the indefinite integral of f .
Thus, means F’(x) = f (x)
( )f x dx
( ) ( )F x f x dx
€
x 2 dx =x 3
3+ C∫
€
x 3
3+ C
€
x 2because the derivative of is
Every antiderivative F of f must be of the form F(x) = G(x) + C, where C is a constant.
Example:
Represents every possible antiderivative of 6x.
Constant of Integration
€
6x dx = 3x 2 + C∫
1
if 11
nn x
x dx C nn
Example:4
3
4
xx dx C
Power Rule for the Indefinite Integral
1 1lnx dx dx x C
x
x xe dx e C
Indefinite Integral of ex and bx
ln
xx b
b dx Cb
Power Rule for the Indefinite Integral
Sum and Difference Rules
f g dx fdx gdx
Example:
2 2x x dx x dx xdx 3 2
3 2
x xC
( ) ( )kf x dx k f x dx ( constant)k
4 43 32 2 2
4 2
x xx dx x dx C C
Constant Multiple Rule
Example:
Integration by Substitution
Method of integration related to chain rule. If u is a function of x, then we can use the formula
/
ff dx du
du dx
Example: Consider the integral:
92 33 5x x dx3 2pick +5, then 3 u x du x dx
10
10
uC
9u du 103 5
10
xC
Sub to get Integrate Back Substitute
Integration by Substitution
Example: Evaluate
3/ 21
10 3/ 2
uC
3/ 225 7
15
xC
25 7x x dxPick u, compute du
Sub in
Sub in
Integrate€
Let u = 5x 2 − 7, du =10x dx
€
1
10⋅10x 5x 2 − 7 dx =
1
10u
12 du∫∫
3ln
dx
x xLet ln then u x xdu dx
3
3ln
dxu du
x x
2
2
uC
2ln
2
xC
Example: Evaluate
Examples on your own:
Find the integral of each:1.) 2.)
3.) 4.)
€
8 dx∫
€
(2x + 6) dx∫
€
(6x 2 −12x + 8) dx∫
€
(9x 2 +12x − 9) dx∫€
F(x) = 8x + c
€
F(x) =2x 2
2+ 6x + c
€
F(x) = x 2 + 6x + c
€
F(x) =6x 3
3−
12x 2
2+ 8x + c
€
F(x) = 2x 3 − 6x 2 + 8x + c
€
F(x) =9x 3
3+
12x 2
2− 9x + c
€
F(x) = 3x 3 + 6x 2 − 9x + c
Find the integral of each:5.) 6.)
7.) 8.)
€
(x − 2)10 dx∫
€
2(2x − 3)4 dx∫
€
(5 − x)6 dx∫
€
(3x −1)4 dx∫€
u = x − 2
du =1 dx
€
u10 du∫
€
F(x) =u11
11+ C
€
F(x) =(x − 2)11
11+ C€
u = 2x − 3
du = 2 dx
€
u4 du∫
€
F(x) =u5
5+ C
€
F(x) =(2x − 3)5
5+ C
€
u = 5 − x
du = −1 dx
€
−u6 du∫
€
−1 u6 du∫
€
F(x) = −u7
7+ C
€
F(x) = −(5 − x)7
7+ C€
u = 3x −1
du = 3 dx
€
1
3u4 du∫
€
1
3u4 du∫
€
F(x) =1
3⋅
u5
5+ C
€
F(x) =(3x −1)5
15+ C
Find the integral of each:9.) 10.)
11.) 12.)
€
1
x + 2dx∫
€
3
x − 4dx∫
€
3x 2 x 3 + 5 dx∫
€
x x 2 − 53 dx∫€
u = x + 2
du =1 dx
€
1
udu∫
€
F(x) = ln u + C
€
F(x) = ln x + 2 + C€
u = x − 4
du =1 dx
€
31
udu∫
€
F(x) = 3ln u + C
€
F(x) = 3ln x − 4 + C
€
u = x 3 + 5
du = 3x 2 dx
€
u du∫
€
u1
2 du∫
€
F(x) =u
32
32
+ C
€
F(x) =2
3x 3 + 5( )
32 + C€
u = x 2 − 5
du = 2x dx
€
u1
3
2du∫
€
F(x) =1
2⋅
u4
3
43
+ C
€
F(x) =3
8x 2 − 5( )
43 + C
Find the integral of each:13.) 14.)
€
10
x 2 dx∫
€
4
x 3 dx∫
€
10x −2 du∫
€
10 x −2 du∫
€
F(x) =10x −1
−1+ C
€
F(x) =−10
x+ C
€
4x −3 du∫
€
4 x −3 du∫
€
F(x) = 4x −2
−2+ C
€
F(x) = −2
x 2 + C
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