Student Study Session
Integration Techniques for the BC Exam Solutions
We have intentionally included more material than can be covered in most Student Study Sessions to
account for groups that are able to answer the questions at a faster rate. Use your own judgment, based
on the group of students, to determine the order and selection of questions to work in the session. Be sure
to include a variety of types of questions (multiple choice, free response, calculator, and non-calculator)
in the time allotted. Do not spend time on the examples provided in the introductory material. These are
provided for later use by students.
Students approach questions requiring the use of u-substitution or parts in a variety of ways. The
solutions to questions 4 and 10 show alternative approaches to each type of question.
1. B (1973 AB30/BC30)
2

2
1
2 1
x4
4


dx     4 x 2  dx   ln x    (ln 2  2)  (ln1  4)  ln 2  2
2
1
x
x 1
x


2. C (1969 AB38)
x2
1  x3
1  x3
1
2
 e x3 dx   3  e (3x ) dx   3 e  C   3e x3  C
3. B (2003 AB8/BC8)
1
1
2
3
2
3
3
 x cos  x  dx  3  3x cos  x  dx  3 sin( x )  c
4. A (1969 AB29)


2
4
1

 2
x
cos x
2
2
dx  ln(sin x) 2  ln1  ln
 ln 

ln
 ln 2


x
sin x
2
2
4
 2 
Alternatively, rewrite the entire integral in terms of u.




2
u  sin x
When x  , u  sin 
; when x  , u  sin  1 .
4
2
2
4
2
du  cos x dx


2
4
1
u 1

u 1 1
cos x
1
dx  2
cos x dx  2 du  ln(u )
u

sin x
2 u
4 sin x
u
2
2
 2
2
2
 ln1  ln
 ln 
 ln 2
  ln
2
2
 2 
Alternatively, integrate in terms of u; replace x before evaluating the definite integral.
u  sin x
du  cos x dx

1
x
2
 2
cos x
1
1
2
2
dx

cos
x
dx

du

ln(
u
)

ln(sin
x
)

ln1

ln

ln
 ln 2

  ln
 sin x
 sin x
u

2
2
2
x


4
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Student Study Session Solutions
Integration Techniques for BC
5. B (1998 BC8 appropriate for AB)
2
2
 sin x cos x dx   sin x  cos x  dx  
1

1

3
 cos x   C . Let x  , 0    cos   C , so C  0 .
3
2
3
2
3
1
1
y (0)   (cos 0)3   .
3
3
6. A (1985 BC40 appropriate for AB)
x
1
u  , du  dx; when x  2, u  1 and when x  4, u  2
2
2
2
 x
1  
2
2
4
 2  dx  2 1  u 2 du  2 1  u du
2 x
1 2u
1 u
7. B (1973 AB38/BC38)
2
Let z  x  c , so 5   f ( x  c) dx  
2c
1c
1
f ( z ) dz .
8. C (1997 AB3)

b
a
b
b
a
a
( f ( x)  5) dx   f ( x) dx  5 1 dx  a  2b  5x a  a  2b  5(b  a)  7b  4a .
b
9. E (1998 AB82)
Since F is an antiderivative of f ,

3
1
x 3
f (2 x) dx 
1
1
F (2 x)   F (6)  F (2)  .
2
2
x 1
10. A (1988 BC16)
v dx
u
x
e2 x
+ 1
e2 x
1
- 2
1 2x
e
0
4
1 2x 1 2x
2x
 x e dx  2 xe  4 e  C
Alternatively, using a non-tabular method with parts,
dv  e2x dx
ux
1
v  e2 x
du  dx
2
1
1 2x
1 2x 1 2x
 2x 
2x
 x e dx  x  2 e    2e dx  2 xe  4 e  C
Copyright © 2013 National Math + Science Initiative®, Dallas, TX. All rights reserved. Visit us online at www.nms.org
Student Study Session Solutions
Integration Techniques for BC
11. E (1993 BC29)
v dx
u
x
sec 2 x
+
sin x
tan x 
1
cos x
 ln | cos x |
0
 x sec
2
x dx  x tan x  ln | cos x | C
12. B (1969 BC42)
u  x2
dv  cos x
du  2 x dx
v  sin x
x
x
2
2
cos x dx  x 2 sin x   2 x sin x dx .
cos x dx  f ( x )   2 x sin x dx
x 2 sin x   2 x sin x dx  f ( x)   2 x sin x dx (using substitution)
x 2 sin x  C  f ( x) .
13. B (1985 BC 21)
u  f ( x)
dv  sin x dx
du  f ( x) dx
v   cos x
 f ( x) sin( x) dx   f ( x) cos x   cos x f ( x) dx
 f ( x) sin( x) dx   f ( x)cos x   3x cos x dx
 f ( x) cos x   cos x f ( x) dx   f ( x) cos x   3 x
2
2
cos x dx (using substitution)
f ( x)  3x 2 , so f ( x)  x3  c .
14. E (2008 BC22)
u  f ( x)
v  g ( x)
du  f ( x) dx
dv  g ( x) dx



1
0
1
0
1
0
1
f ( x) g ( x) dx  f ( x) g ( x) 10   g ( x) f ( x)dx
0
f ( x) g ( x) dx  f (1) g (1)  f (0) g (0)  5 (using substitution)
f ( x) g ( x) dx 4(3)  2(4)  5  15
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Student Study Session Solutions
Integration Techniques for BC
15. A (1985 BC12)
1  A( x  2)  B( x  1)
1
1
Let x  2 : 1  B(2  1), so B   ; Let x  1: 3 A  1, so A  .
3
3
1 
 1
 
 3
1
1
1 x 1
  x  1  x 32  dx  3 ln | x  1|  3 ln | x  2 | c  3 ln x  2  C




16. A (2008 BC19)
7 x  A( x  2)  B(2 x  3)
3
7 
7  2 
3
2
 2.
Let x  : A     3 ; Let. x  2 : B 
3
2
2(2)  3
2
2
2 
3
 3
  2 x  3  x  2  dx  2 ln | 2 x  3 | 2 ln | x  2 | C
17. E (1973 BC36)
b 2( x  1)
( x  1)
1
1
1
dx  lim  2
dx  lim ln | x 2  2 x  3 |b0  lim  ln | b 2  2b  3 |  ln | 3 |   
0 x  2x  3
2 b1 0 x  2 x  3
2 b1
2 b1
so this limit diverges.

1
2
18. A (1988 BC7)
b
lim 
b
b  2
 1
 1 1 1
x dx  lim     lim      .
b 
 x  2 b  b 2  2
2
19. E (1993 BC11)
lim 

b  4
9  x 
2

1
3
b
2
3
3
3

2 3

2
x
dx

lim
(9

x
)
 lim  3 9  b 2  3 9  4 2  . This limit diverges.
 
b  2
b  2
2


4
20. C (1998 AB88)
Using a calculator, F (9)  F (1)  
9
1
(ln x)3
dx  5.827 .
x
21. D (1997 BC89)
Using a calculator, f (4)  f (1)  
4
1
x2
dx  0.376 .
1  x5
Copyright © 2013 National Math + Science Initiative®, Dallas, TX. All rights reserved. Visit us online at www.nms.org
Student Study Session Solutions
Integration Techniques for BC
22. 1976 BC4
(a) Let u  x 2
dv  e5x dx
1
du  2 xdx v  e5 x
5
1
2
Then  x 2e5 x dx  x 2e5 x   xe5 x dx
5
5
Let u  x dv  e5x dx
1
du  dx v  e5 x
5
1
2 1
1

 x 2e5 x   xe5 x   e5 x dx 
5
5 5
5

1
2
2 5x
 x 2e5 x  xe5 x 
e C
5
25
125
(b) Let u  x n
dv  ekx dx
1
du  nx n1dx v  e kx
k
1 n kx n n1 kx
n kx
 x e dx  k x e  k  x e dx
23. 2004B BC5d

1
dx  lim(2 b  4)  
4
b  4
b 
x
This limit is not finite, so the integral is
divergent.
(d)  g ( x) dx  lim 

b
4
g ( x) dx
b4

b
1 b 1
2 b 4
dx 

b4 4 x
b4
1:
4

b
4
g ( x) dx  2 b  4
1: indicates integral diverges
1 b
2 b 4
g ( x) dx 
1:

b4 4
b4
1: finite limit as b  
2 b 4
0
b  b  4
lim
Copyright © 2013 National Math + Science Initiative®, Dallas, TX. All rights reserved. Visit us online at www.nms.org
Student Study Session Solutions
Integration Techniques for BC
24. 1971 BC5
Let I   xe  x dx
No points shown.
Let u  x dv  e x dx
du  dx v  e x
 I   xe  x   e  x dx
  xe  x  e  x


0
b
xe x dx  lim  xe x dx
b 0
 lim   xe x  e x 
b
b
0
 lim  be  e  1
b
b
b
 b 1

 lim   b  b  1
b 
 e e

1
 Integral converges to 1.
Copyright © 2013 National Math + Science Initiative®, Dallas, TX. All rights reserved. Visit us online at www.nms.org