Download 4 .3 Integration by Substitution

Section 4.3
n 1
u
n
u
 du  n  1  C, n   1
1.  (x2  1)9 2x dx
Let u = x 2 +1. Then du = 2x dx and the integral becomes
u10
 (x  1) 2x dx   u du  10  C
2
9
9
And reversing the substitution yields:
(x 2  1)10
C
10
2.  e 5x4 dx
x5
∫ e u du = e u + C
Let u = x 5 . Then du = 5x4 dx and the integral becomes
u
u
e
du

e
C

And reversing the substitution yields:
e C
x5
3.  (x  16) x dx
4
5
3
n 1
u
n
u
 du  n  1  C, n   1
Let u = x 4 - 16. Then du = 4x3 dx and we need to make an adjustment for
the 4.
1
4
5
3
(x

16)
4x
dx

4
And then the integral becomes
1 5
1 u6
(x4  16)6
u du 
C
C

4
4 6
24
4.  e3x dx
∫ e u du = e u + C
Let u = 3x. Then du = 3 dx and we need to make an adjustment for the 3.
1 3x
e 3dx
3
And then the integral becomes
1 u
1 u
1 3x
e du  e  C  e  C
3
3
3
1
5. 
dx
1  5x

1
du  ln u  C
u
Let u = 1 + 5x. Then du = 5 dx and we need to make an adjustment for the 5.
1
5
1
 1  5x 5 dx
And then the integral becomes
1
5
1
1
du

ln
u

C

ln 1  5x  C
u
5
6.

4
x4  16 x 3 dx
un  1
 u du  n  1  C, n   1
n
Let u = x 4 + 16. Then du = 4 x 3 dx and we need to make an adjustment for the 4.
1 4 4
3
x

16
4x
dx

4
And then the integral becomes
1 14
1 4 54
1 4
5 4
u
du

u

C

(x

16)
C
4
4 5
5
7.  (2x2  4x)5 (x  1) dx
n 1
u
n
u
 du  n  1  C, n   1
Let u = 2x 2 + 4x. Then du = 4x +4 dx = 4(x + 1) dx and we need to make an
adjustment for the 4.
1
2
5
(2x

4x)
4(x  1) dx

4
And then the integral becomes
1 5
11 6
1
2
6
u
du

u

C

(2x

4x)
C

4
46
24
x3  x2
8.  4
dx
3x  4x3

1
du  ln u  C
u
Let u = 3x 4 + 4x 3 . Then du = 12x 3 + 12x 2 dx = 12(x 3 + x 2) dx and we need to
make an adjustment for the 12.
1 12(x3  x2 )
dx
4
3

12 3x  4x
And then the integral becomes
1 1
1
1
4
3
du

ln
u

C

ln
3x

4x
C
12  u
12
12
x
9. 
dx
2
1x

1
du  ln u  C
u
Let u = 1 – x 2. Then du = - 2x dx and we need to make an adjustment for the - 2.

1
2
2x
 1  x2 dx
And then the integral becomes

1
2
1
1
1
2
du


ln
u

C


ln
1

x
C
u
2
2
e2x
10.  2x
dx
e 1

1
du  ln u  C
u
Let u = e 2x + 1. Then du = 2e 2x dx and we need to make an adjustment for the 2.
1
2
2e2x
 e2x  1 dx
And then the integral becomes
1
2
1
1
1
2x
du

ln
u

C

ln
e
1 C
u
2
2
11.
ln x
 x dx
Let u = ln x. Then du = 1/x dx.
And then the integral becomes
u2
(ln x)2
 u du  2  C  2  C
12.  (x  1)x2 dx
This one is a little different. We need to do a little algebra first. Distribute the x 2.
x4
x3
 (x  1) x dx   (x  x ) dx  4  3  C
2
3
2
13. BUSINESS: Cost – The weekly marginal cost of producing shoes is given by
C’ (x) = 12 + 500/(x + 1) where C (x) is the cost in dollars. If the fixed cost
are $2,000 per week, find the cost function.
Step 1: Integrate C’ to find C.
Let u = x + 1 then du = dx and the integral becomes
 12 
500
dx   12dx 
x1
500
 u du  12x  500 ln u  C  12x  500 ln x  1  C
C(x)  12x  500 ln x  1  C
Step 2: Find C. The fixed cost are 2000 or C(x) = 2000 when x = 0.
12(0)  500 ln 0  1  C  2000
0  0  C  2000 and C  2000
C(x)  12x  500 ln x  1  2000
14. BUSINESS: Price-demand – The marginal price p’ (x) at x boxes of a certain cereal
per week is given by p’ (x) = - 0.015e -.01x . Find the price equation if the weekly demand
is 50 boxes when the price is $4.35.
Step 1: Integrate p’ to find p.
Let u = - 0.01x then du = - 0.01 and the integral becomes
 0.01x
 0.01x
  0.015e dx   0.015 e dx 
 0.015  0.01x
 0.015 u
e
(

0.01)dx

e du
 0.01 
 0.01 
And integration yields
p(x) 
0.015 u
e  C  1.5e 0.01x  C
0.01
Step 2: Find p (x). At a price of $4.35, 50 boxes are sold, so
4.35  1.5e 0.01(5)  C  .91  C
3.44  4.35  .91  C
p(x)  1.5e 0.01x  3.44
15. MEDICINE – The rate of healing for a skin wound (in square centimeters per day)
is approximated by A’ (t) = - 0.9e -0.1t . If the initial wound has an area of 8 square
centimeters, what will its area be after t days? After 5 days?
Step 1: Integrate A’ to find A.
A   0.9 e 0.1t dt
Let u = - 0.1 t then du = - 0.1 dt and the integral becomes
A   0.9  e  0.1t dt 
 0.9  0.1t
 0.9 u
e
(

0.1
)
dt

e du


 0.1
 0.1
A  9e 0.1t  C
Step 2: Find C. The size of an initial wound is 8 when x = 0.
8  9e 0.1(0)  C and C   1
Step 3. Find A (5)
A(5)  9e 0.1(5)  1  4.46 cm2
A  9e 0.1t  1
16. POLLUTION - A contaminated lake is treated with a bactericide. The rate of
increase in harmful bacteria t days after the treatment is given by
dN
2000t

dt
1  t2
for 0  t  10
Where N (t) is the number of bacteria per milliliter of water.
a. Find the minimum value of dN/dt.
b. If the initial count was 5,000 bacteria per milliliter, find N (t).
c. Find the bacteria count after 10 days.
a. Use your calculator to minimize
the given function.
The minimum value is – 1000.
b. Integrate dN/dt ’ to find N.
N (t)   
N (t)  
2000t
dt
1  t2
Let u = 1 + t 2 then du = 2t dt and the integral becomes
2000
2t
1
dt


1000
 u du   1000ln u  C
2  1  t2
N(t)   1000 ln 1  t 2  C
CONTINUED
N(t)   1000 ln 1  t 2  C
Find C. The initial count was 5000 when t = 0.
5000   1000 ln 1  02  C or 5000  C
N (t)   1000 ln 1  t 2  5000
c. Find N (10).
N (10)   1000 ln 1  102  5000  385
10 days after treatment the bacteria count will be 385 bacteria per milliliter of water.