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MATH1111 Quizzes

Integration by Substitution Quiz
Web resources available Questions

This quiz tests the work covered in lecture on integration by substitution and corresponds to Section 7.1 of the textbook Calculus: Single and Multivariable (Hughes-Hallett, Gleason, McCallum et al.).

There are more web quizzes at Wiley, select Section 1.

The MATH1011 Quiz 11 should also be appropriate to try. Only questions 4, 5, 8, 9 and 10 involve integration by substitution. The other questions would be good revision. These and other sources use u rather than w as the letter for the substituted function.

Visual calculus again has good examples at http://archives.math.utk.edu/visual.calculus/4/substitutions.3/ and there are plenty of exercises at http://archives.math.utk.edu/visual.calculus/4/substitutions.1/index.html with hints. The visual calculus quiz has some very hard questions.

If you are looking for another explanation then try the pdf file at http://www.mathcentre.ac.uk/students.php/all_subjects/integration/by_substitution/resources/79.

Finally the Maths Learning Centre Booklet Introduction to Integration is a good source of explanations and exercises.

Which of the following substitutions are correct? (Zero or more options can be correct)
a)
tet2+4 dt =ewdw where w = t2 + 4.
b)
3x2 + 4x + 1 x3 + 2x2 + x 3dx = 1 wdw where w = x3 + 2x2 + x 3.
c)
cosx 2x dx =coswdw where w = x.
d)
1 x2 + 4dx = 1 wdw where w = x2 + 4.
e)
(z2 + 2z)(z3 + 3z2 4)4dz =w4 3 dw where w = z3 + 3z2 4.

There is at least one mistake.
For example, choice (a) should be False.
The constant is not correct
xex2+4 dx =1 2ewdw where w = x2 + 4.
There is at least one mistake.
For example, choice (b) should be True.
w = x3 + 2x2 + x 3 so dw = (3x2 + 4x + 1)dx as required.
There is at least one mistake.
For example, choice (c) should be True.
w = x = x1 2 so dw = 1 2x1 2 dw = 1 2xdx as required.
There is at least one mistake.
For example, choice (d) should be False.
This integral cannot be solved using this method of substitution.
There is at least one mistake.
For example, choice (e) should be True.
w = z3 + 3z2 4 and dw = (3z2 + 6z)dx so (z2 + 2z)dx = 1 3dw as required
Correct!
  1. False The constant is not correct
    xex2+4 dx =1 2ewdw where w = x2 + 4.
  2. True w = x3 + 2x2 + x 3 so dw = (3x2 + 4x + 1)dx as required.
  3. True w = x = x1 2 so dw = 1 2x1 2 dw = 1 2xdx as required.
  4. False This integral cannot be solved using this method of substitution.
  5. True w = z3 + 3z2 4 and dw = (3z2 + 6z)dx so (z2 + 2z)dx = 1 3dw as required
Find the indefinite integral x2(x3 + 4)2dx. Exactly one option must be correct)
a)
x2(x3 + 4)2dx = (x3 + 4)3 + C
b)
x2(x3 + 4)2dx = 2(x3 + 4)3 3 + C
c)
x2(x3 + 4)2dx = (x3 + 4)3 9 + C
d)
x2(x3 + 4)2dx = (x3 + 4)2 6 + C

Choice (a) is incorrect
Try again, you have not dealt with the constants correctly.
Choice (b) is incorrect
Try again, you have not dealt with the constants correctly.
Choice (c) is correct!
Let w = x3 + 4 then dw = 3x2dx and x2dx = 1 3dw.

x2(x3 + 4)2dx =1 3w2dw = 1 9w3 + C = (x3 + 4)3 9 + C.
Choice (d) is incorrect
Try again, you have not integrated correctly.
Find the indefinite integral (x3 + 2)cos(x4 + 8x)dx. Exactly one option must be correct)
a)
(x3 + 2)cos(x4 + 8x)dx = sin(x4 + 8x) 4 + C
b)
(x3 + 2)cos(x4 + 8x)dx = sin(x4 + 8x) 4 + C
c)
(x3 + 2)cos(x4 + 8x)dx = cosw 4 + C
d)
This integral cannot be integrated using substitution.

Choice (a) is incorrect
Try again, you have not integrated cosw correctly.
Choice (b) is correct!
Let w = x4 + 8x then dw = (4x3 + 8)dx = 4(x3 + 2)dx and (x3 + 2)dx = 1 4dw.
(x3 + 2)cos(x4 + 8x)dx = 1 4 coswdw = 1 4 sinw = 1 4sin(x4 + 8x) + C
Choice (c) is incorrect
Try again, you have not finished the integral.
Choice (d) is incorrect
Try again, let w = x4 + 8x.
Which of the following integrals have been evaluated correctly? (Zero or more options can be correct)
a)
02x2(x3 + 1)12 =19w12 3 dw = 2 9w32 19 = 52 9
b)
0πθsin(θ2)dθ =0πw 2 dw = w2 4 0π = π2 4
c)
12 dx (3 5x)2 = 1 572du u2 = 1 5u 72 = 1 14
d)
043x + 4dx = 1 3416wdw = 1 3 2 3w32 416 = 112 9
e)
01t2et3 dt =01 ew 3 dw = ew 3 01 = 1 3 1 3e

There is at least one mistake.
For example, choice (a) should be True.
There is at least one mistake.
For example, choice (b) should be False.
There are a variety of errors. The limits of integration have not been changed and the integration has not been performed correctly. 0πθsin(θ2)dθ =0π2 sinw 2 dw = cosw 2 |0π2 = 1 2(cosπ2 1)
There is at least one mistake.
For example, choice (c) should be False.
The answer is correct but the steps are incorrect. The limits of integration must stay in the correct order.
12 dx (3 5x)2 = 1 527du u2 = 1 5u27 = 1 14
There is at least one mistake.
For example, choice (d) should be True.
There is at least one mistake.
For example, choice (e) should be True.
Correct!
  1. True
  2. False There are a variety of errors. The limits of integration have not been changed and the integration has not been performed correctly. 0πθsin(θ2)dθ =0π2 sinw 2 dw = cosw 2 |0π2 = 1 2(cosπ2 1)
  3. False The answer is correct but the steps are incorrect. The limits of integration must stay in the correct order.
    12 dx (3 5x)2 = 1 527du u2 = 1 5u27 = 1 14
  4. True
  5. True