Student Study Session
Integration Techniques for AB 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.
Students approach u-substitutions questions in a variety of ways. The solution to question 1
demonstrates three different approaches.
Multiple Choice Questions Solutions
1. E (1973 AB27)
1
2
0

2x
1  x2
1
2
0
dx    (1  x )
2

1
2
1
1 x 2
2 2
(2 x) dx  2(1  x )
x 0
2


2
1

 2 1     1   0     3  2


2


Alternatively, rewrite entire intregral in terms of u
2
when x  0 , u  1  02  1 and when x 
u  1  x2
du  2 x dx
du  2 x dx
1
2
0

1
2x
1  x2
dx    2 (1  x 2 )

1
2
0
3
1
1
, u  1   
2
4
2
(2 x) dx
3
1 u 4
2
 3

 2 
 1    3  2
 4

u 1
Alternatively, integrate in terms of u; replace x before evaluating definite integral
u  1  x2
du  2 xdx
du  2 x dx
3
4
u 1
 
1
2
0

u
(u )
1
2x
1  x2

1
2
dx    2 (1  x 2 )
0
   (u )

1
2
du  2u

1
2
(2 x) dx
1
2
1
1 x 2
2 2
du  2u  2(1  x )
2. B (1985 BC3 appropriate for AB)
2 x 1
1 2 (2 x  2)
1
dx

dx  ln x 2  2 x
2
1 x 2  2 x

1
2 x  2x
2
x 0
x2
x 1

2


2
1
 2  1     1   0     3  2


2


1
(ln 8  ln 3)
2
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Integration Techniques for the AB exam
Student Study Session
3. 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


4. E (1998 AB7/BC7)
e

e
1
e
1
x2 1
3
1

1
 1
 1
dx =   x   dx   x 2  ln x  =  e 2  1    0   e 2 
1
2
x
x
 2
 2

2
1  2
5. C (1997 BC1 appropriate for AB)

1
0
1
 32

2 52 2 23
2
x ( x  1) dx    x  x  dx  x  x
0
5
3


x 1
1

x 0
16
15
6. C (1969 AB38)
x2
1  x3
1  x3
1
2
 e x3 dx   3  e (3x ) dx   3 e  C   3e x3  C
7. B (2008 AB4)
1
1
1
1
 sin(2 x) dx   cos(2 x) dx  2  2sin(2 x) dx  2  2 cos(2 x) dx   2 cos(2 x)  2 sin(2 x)  C
8. B (2003 AB8/BC8)
1
1
2
3
2
3
3
 x cos  x  dx  3  3x cos  x  dx  3 sin( x )  C
9. A (1969 AB29)


x
cos x
2
2
2
4 sin x dx  ln(sin x) x42  ln1  ln 2  ln 2  ln 2
10. B (1998 BC8 appropriate for AB)
 sin x cos
2
x dx   sin x  cos x  2 dx  
1
3
 cos x   C .
3
1

Let x  , 0    cos   C , so C  0 .
2
3
2
1
1
y (0)   (cos 0)3   .
3
3

3
11. 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
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Integration Techniques for the AB exam
Student Study Session
12. C (1988 AB10)
xk
1
2
18  kx  x3
 k 3  k 3  27, k  3
3 x 0 3
2
13. B (1973 AB38/BC38)
2
Let z  x  c , so 5   f ( x  c) dx  
2c
1c
1
f ( z ) dz .
14. 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
15. A (1998 AB11/BC11)
Since f is linear, its second derivative is zero. The integral gives the area of a rectangle with
zero height and width b  a. This area is zero.
16. A (1998 AB20)
xk
1 3
1 3
1 3
2
3
3 x dx  3 x x3  3 (k  (3)  3 (k  27)  0 only when k  3.
k
17. E (1998 AB82)

Since F is an antiderivative of f ,
3
1
x 3
1
1
f (2 x) dx  F (2 x)   F (6)  F (2)  .
2
2
x 1
18. D (1997 AB90)
1
I. f ( x)  (2sin x)(cos x)  sin x cos x
2
1
II. f ( x)  (2 cos x)( sin x)   sin x cos x
2
1
III. f ( x)   ( sin(2 x))(2)
4
1
 sin(2 x)
2
1
 (2sin x cos x)  sin x cos x
2
The derivative of the first and last functions are equal to f ( x) .
19. C (2008 BC86 appropriate for AB)
dy
 2 x and y  3 when x  2 . y  x 2  C ; 3  4  C , so C  1 ; therefore, y  x 2  1 and
dx
f (3)  32  1  8
20. C (1998 AB88)
Using a calculator, F (9)  F (1)  
9
1
(ln t )3
dt  5.827 .
t
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Integration Techniques for the AB exam
Student Study Session
Free Response Solutions
21. 1993 AB5
if 0  x  1
 x
(a) f   x   
2  x if 1  x  2
OR
f   x   1  x  1 if 0  x  2
 x2
if 0  x  1
 2  C1
(b) f  x   
2
2 x  x  C if 1  x  2
2

2
1
1
0  f 1   C1  C1  
2
2
1
3
0  f 1  2   C2  C2  
2
2
2
 x 1
if 0  x  1
 2  2
f  x  
2
2 x  x  3 if 1  x  2

2 2
OR
1
f  x   x   x  1 x  1  1
2
(c)
2
1: x if x  1
1: 2  x if x  1
OR
2: 1  x  1
1: Antiderivative for x  1 with or
4 1: Antiderivative for x  1 without
C’s
2: Constants
OR
3: Antidifferentiates absolute value
function
2 If incorrect on one branch
1: Constant
4
0 of 4 if not a split function or
absolute value function
NOTE: No penalty for domain error
in Part (b) if consistent with
domain in Part (a)
Consistency with given f '
1: Increasing and continuous on
 0, 2 .
1: Concave up on  0, 1 and concave
down on 1, 2  .
3 1:
1
f 1  0 , lim f  x    ,
x0
2
1
lim f  x  
x  2
2
or if f 1  0 not used in Part (b),
then
 lim f  x   lim  f  x    1 .
 x2

x 0
If student goes outside domain,
loses 3rd point.
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Integration Techniques for the AB exam
Student Study Session
22. 1989 BC1 (appropriate for AB)
(a)
f   x   3x 2  8 x  c
1: f   x  with c
f  0  3
c3
1: Sets f   0   3
2: 
5
1: Finds c
1: uses his/her f  to find f with constant
f  x   x3  4 x2  3x  d
d  2
f  x   x3  4x2  3x  2
(b)
1: uses  0,  2 to find d
1
1
x3  4 x 2  3x  2  dx



1
1   1
2 : integer and other his/her f  x 

1

3 : Set up 2 : Incorrect
ba

4
1: Incorrect limits
1: antidifferentiation and evaluation
1
1 1
4
3

  x 4  x3  x 2  2 x 
2 4
3
2
 1
1  1 4 3
 1 4 3

    2       2 

2  4 3 2
 4 3 2

2

3

23. 1988 AB6
a b  6
f   x   2ax  b
4
2a  b  18
a  12 , b  6
and f   x   12x2  6x
f  x    12 x 2  6 x  dx
 4 x3  3x 2  C
  4x
2
1
3
 3x2  C  dx  x4  x3  Cx 1
2
 16  8  2C   1  1  C 
 8  C  18 or
C  10
1: for a  b  6
1: for f   x 
1: for 2a  b  18
1: for a  12 , b  6
for an antiderivative of f 
for including C
for antiderivative of f
for evaluating integral and
equating it to 18
1: for finding value of C
1:
1:
5 1:
1:
 f  x   4 x3  3x2  10
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Integration Techniques for the AB exam
Student Study Session
24. 1985 AB4/BC3
1 4
(a) Ave a   14 x 2 dx
3 1
2: for correct integral expression
3 1: for antidifferentiation and evaluation
4
14 x3
14



 64  1
3 31
9
 98
1 k 2
x
k sin
dx

0
k
2k
k
2k 2
x

cos

2k 0
(b) Aveb 

2k 2
2k


2
 0  1 
2k 2
2:
2:
6
1:
1:
for correct integral expression
for antidifferentiation
for setting Avea  Aveb
for solving for k

 98
k 2  49 2
k  7
Copyright © 2013 National Math + Science Initiative®, Dallas, TX. All rights reserved. Visit us online at www.nms.org