College Mathematics: Learning Worksheets Chapter 13
The producers’ surplus will be
0
100
[()]
[25 (10 0.15 )]
x
PS p S x dx
PS x dx
College Mathematics: Learning Worksheets Chapter 13
Name ________________________________ Date ______________ Class ____________
Goal: To solve integration problems using the integration by parts formula
Section 13-3 Integration by Parts
Formula: Integration by Parts
udv uv vdu
Summary: Integration by Parts: Selection of u and dv
For ,udv uv vdu
2. It must be possible to integrate dv (preferably by using standard formulas or
simple substitution).
4. For integrals involving ,
pax
x
etry
p
ux and ax
dv e dx
5. For integrals involving (ln ) ,
pq
x
xtry
College Mathematics: Learning Worksheets Chapter 13
In Problems 1 and 2, integrate by parts. Assume that x > 0 whenever the natural logarithm
function is involved.
1. 12x
x
edx
∫
Let uxand 12
x
dv e dx=, then du dxand 12
1
12 .
x
ve= Substituting these terms into
the integration by parts formula yields
x
x
x
2. 4ln
x
xdx
∫
x
5.vx= Substituting these terms into
the integration by parts formula yields
udv uv vdu
=−
∫∫
3. 8
1ln 7
x
dx
∫
Use integration by parts by letting ln 7ux=and ,dv dxthen 1
x
du dxand .vx
Substitute these terms into the integration by parts formula yields
8
88
1
11
1
ln 7 (ln 7 )( ) ( )
x
xdx x x x dx
=−
∫∫
4. 9x
x
edx
Use integration by parts by letting 9uxand
x
dv e dx, then 9du dxand .
x
ve
Substituting these terms into the integration by parts formula yields
x
x
5. 3
5
3x
x
edx
3.
x
ve Substituting these terms into the integration by parts formula yields
53 2
xxx
x
College Mathematics: Learning Worksheets Chapter 13
390
6.
4
4
43
3
x
x
edx
ex
Use substitution by letting 43
x
ue xand 4
43.
x
du eSubstitute these terms into
the original equation yields
4
4
43 1
x
x
edx du
7. 2
22
0
x
x
edx
Use substitution by letting 2
2ux and 4.du x Substitute these terms into the
original equation yields
2
22
2
00
1
4
x
xu
x
xe dx e du
8. 8
(7)(3)
x
xdx
Use integration by parts by letting 3uxand 8
(7)dv x dx , then du dxand
9
1
9(7).vx Substituting these terms into the integration by parts formula yields
899
11
99
(7)(3) (3)((7)) (7)
x
xdxx x x dx
x
College Mathematics: Learning Worksheets Chapter 13
391
Name ________________________________ Date ______________ Class ____________
Goal: To use the Trapezoidal Rule and Simpson’s Rule to approximate definite integrals.
Use integration tables to solve indefinite integrals
Section 13-4 Other Integration Methods
Trapezoidal Rule:
Let f be a function defined on an interval [a, b]. Partition [a, b] into n
subintervals of equal length ()/
x
ba n with endpoints
012 n
ax x x x b
Then
012
[( ) 2( ) 2( ) ( )] /2
nn
Tfx fx fx fxx
is an approximation of ( ) .
b
a
f
xdx
Simpson’s Rule:
Let f be a function defined on an interval [a, b]. Partition [a, b] into 2n
subintervals of equal length ()/
x
ba n with endpoints
012 n
ax x x x b
Then
f
Process: Using the integration tables
1. Determine which category has the form of the given integrand (integrals involving
trigonometric functions, integrals involving ln ,uetc.)
2. Examine the integrals within the category to see if one represents exactly the
integrand in the given problem.
a) If an exact match is found, use the formula to find the indefinite integral.
b) If an exact match is not found, use substitution to change the integral into one
392
1.
2
916
xdx
x
∫−
Examining the form of the integral, we should look at formulas 40 to 45, which
involve 22
.ua Based on this, we let 22
9ux=and 216a=so that 222
916 .
x
ua−= −
Then we have 3,ux=22
1
9,
x
u=and 3.du dx= Using formula 44 and the substitutions, the
integral will be written as follows:
2
1
29
222
2
22
1
3
916 4
1
27 4
u
xdx du
xu
udu
u
=
∫∫
−−
=∫−
2. 5
1
1
(6 5 ) dx
xx
∫+
Examining the form of the integral, we should look at formulas 3 to 12, which
involve .abu Based on this, we let u = x. Therefore, we use formula 9 with 6a=and
5.b= Use this information to rewrite and solve the integral as follows:
55
11
5
11
(6 5 ) ( )
x
x
x
dx du
xx uabu
=
=
=
=
∫∫
++
College Mathematics: Learning Worksheets Chapter 13
3.
2
2
16 36
16
xdx
x
Examining the form of the integral, we should look at formulas 32 to 39, which
involve 22
.ua Based on this, we let 22
16uxand 222
16 36 .
x
ua Then we
have 4uxand 4.du dx Using formula 35 and the substitutions, the integral can be written
as follows:
222
22
16 36 1 6
4
xu
dx du
4. 1
0
35
47
x
dx
x
Examining the form of the integral, we should look at formulas 15 to 20, which
involve abuand .cdu Based on this, we let u = x. Therefore, we use formula 20 with
3,a5,b4,cand 7.d Use this information to rewrite and solve the integral as
follows:
11
00
35
47
x
x
xabu
dx du
xcdu
5. 2
49 16 dx
x
Examining the form of the integral, we should look at formulas 13 and 14 which
involve 22
.au Based on this, we let 22
16uxand 222
49 16 .
x
au Then we have
7,a4,uxand 4.du dx Using formula 14 and the substitutions, the integral can be
written as follows:
222
111
4
49 16 7
dx du
xu
6. Use the trapezoidal rule with n = 4 to approximate
83
0
1.
x
dx
Partition [0, 8] into 4 equal subintervals of width (8 – 0) / 4 = 2. The endpoints are
0123 4
0, 2, 4, 6, and 8.xxxx x We calculate the value of the function at each
endpoint:
x f(x)
0 1
2 3
6 14.7309
By the trapezoidal rule:
[(0) 2(2) 2(4) 2(6) (8)] /2
Tf f f f f x
College Mathematics: Learning Worksheets Chapter 13
395
7. Use Simpson’s rule with n = 3 to approximate
7
2
1
1.
1dx
x
Partition [1, 7] into 6 equal subintervals of width (7 – 1) / 6 = 1. The endpoints are
01 2 3 4 5 6
1, 2, 3, 4, 5, 6, and 7.xxx x x x x We calculate the value of the function
at each endpoint:
x f(x)
1 .5
2 .2
3 .1
College Mathematics: Learning Worksheets Chapter 13
396