Methods of Integration (College Board AP® Calculus AB): Exam Questions

1 hour37 questions
1
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1 mark

3x2sin(x3+4) dx=

  • cos(x3+4)+C

  • cos(x3+4)+C

  • x3cos(x3+4)+C

  • x3cos(x3+4)+C

2
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1 mark

4x32x23x+7 dx=

  • ln|12x23x+7|+C

  • ln|4x32x23x+7|+C

  • ln|2x23x+7|+C

  • 12x218x4x39x2+42x

3
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1 mark

01(4x+6)(x2+3x1)3 dx=

  • 20

  • 30

  • 40

  • 50

4
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1 mark

13x2x3+1 dx=

  • 13ln3

  • 13ln14

  • ln3

  • ln14

5
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1 mark

(cos(3x)+sin(3x))dx=

  • 13sin(3x)13cos(3x)+C

  • 13sin(3x)+13cos(3x)+C

  • 3sin(3x)3cos(3x)+C

  • 3sin(3x)+3cos(3x)+C

6
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1 mark

02e3xdx=

  • e63

  • 3e6

  • e61

  • 1e63

7
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1 mark

Using the substitution u=3x+2, 28(3x+2)7dx is equivalent to

  • 1302u7du

  • 13826u7du

  • 02u7du

  • 826u7du

8
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1 mark

01(2x1)4dx=

  • 0

  • 15

  • 25

  • 45

1
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1 mark

Using the substitution u=x, 49cosxx dx is equal to which of the following?

  • 1223cosu du

  • 223cosu du

  • 49cosu du

  • 249cosu du

2
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1 mark

xx1 dx=

  • xln|x|+C

  • xln|x1|+C

  • x+ln|x1|+C

  • x2x22x+c

3
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1 mark

π2πsinx1cosxdx=

  • 2(2+1)

  • 2(21)

  • 2(21)

  • 2(2+1)

4
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1 mark

1x26x+10dx=

  • arcsin(x3)+C

  • arctan(x3)+C

  • ln|x26x+10|+C

  • 1(x26x+10)2+C

5
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1 mark

x3+2x2+2x+2x+1dx=

  • 3x3+8x2+12x+246x+12+C

  • 13x3+12x2+2x+C

  • 13x3+12x2+x+ln|x+1|+C

  • (14x4+23x3+x2+2x)·ln|x+1|+C

6
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1 mark

x3x49dx=

  • 18(x49)2+C

  • 14ln|x49|+C

  • 4ln|x49|+C

  • 13arctan(x23)+C

7
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1 mark

x3sin(x4)dx=

  • 14cos(x4)+C

  • 14cos(x4)+C

  • x34cos(x4)+C

  • x44cos(x55)+C

8
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1 mark

x2dx5x3+1=

  • 130(5x3+1)12+C

  • 215(5x3+1)12+C

  • 23x3(5x3+1)12+C

  • 245(5x3+1)32+C

1
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1 mark

x125x2 dx=

  • 15arcsin(5x)+C

  • x5arcsin(5x)+C

  • 125125x2+C

  • 125ln125x2+C

2
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1 mark

Let f be a function such that 312f(3x)dx=9. Which of the following must be true?

  • 14f(t) dt=3

  • 14f(t) dt=27

  • 936f(t) dt=3

  • 936f(t) dt=27

3
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1 mark

If f is a continuous function and if F'(x)=f(x) for all real numbers x, then 496xf(x2)dx=

  • F(3)6F(2)6

  • 3F(3)3F(2)

  • F(81)6F(16)6

  • 3F(81)3F(16)

4
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1 mark

(1cosx 0xcost dt) dx=

  • cosx1cosx+C

  • (cosx1)1cosx+C

  • 32(1cosx)+C

  • 23(1cosx)32+C

5
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1 mark

13+2xx2dx=

  • arcsin(x12)+C

  • 2arcsin(x12)+C

  • arcsin(x1)+C

  • 2arcsin(x1)+C

6
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1 mark

x42x3+x27x2dx=

  • 14x423x3+12x27ln|x2|+C

  • 14x4+12x2+2x3ln|x2|+C

  • x4+2x2+8x12ln|x2|+C

  • 3x48x3+6x284ln|x2|+C