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Integration by Substitution Quiz

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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.


 

Question 1

 
 
Which of the following substitutions are correct?
a) ∫ ∫ t2+4 w te dt = e dw where w = t2 + 4.
b) ∫ ∫ --3x2 +-4x-+-1 dx = 1-dw x3 + 2x2 + x- 3 w where w = x3 + 2x2 + x - 3.
c) ∫ √-- ∫ cos√-x-dx = cosw dw 2 x where w = √x-.
d) ∫ ∫ --1---dx = 1-dw x2 + 4 w where w = x2 +4 .
e) ∫ ∫ (z2 + 2z)(z3 + 3z2 - 4)4 dz = w4dw 3 where w = z3 + 3z2 - 4.

 

There is at least one mistake.
For example, choice (a) should be false.
The constant is not correct
∫ ∫ xex2+4dx = 1 ewdw 2 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-= x 12 so dw = 1x - 12 dw =-√1-dx 2 2 x 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 = 13 dw as required
Your answers are correct
  1. False. The constant is not correct
    ∫ ∫ xex2+4dx = 1 ewdw 2 where w = x2 + 4.
  2. True. w = x3 + 2x2 + x - 3 so dw = (3x2 +4x + 1)dx as required.
  3. True. w = √x-= x 12 so dw = 1x - 12 dw =-√1-dx 2 2 x 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 = 13 dw as required
 

Question 2

 
 
Find the indefinite integral ∫ x2(x3 + 4)2dx.
a) ∫ x2(x3 + 4)2dx = (x3 + 4)3 + C   b) ∫ 2(x3 + 4)3 x2(x3 + 4)2dx =---------+ C 3
c) ∫ 2 3 2 (x3 + 4)3 x (x + 4) dx = ---9----+ C   d) ∫ 2 3 2 (x3 +-4)2 x (x +4) dx = 6 + C

 

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

∫ ∫ 2 3 2 1 2 1 3 (x3 +-4)3 x (x +4) dx = 3 w dw = 9w + C = 9 +C .
Not correct. Choice (d) is false.
Try again, you have not integrated correctly.
 

Question 3

 
 
Find the indefinite integral ∫ 3 4 (x + 2)cos (x + 8x)dx.
a) ∫ 3 4 --sin(x4-+8x-) (x + 2)cos(x + 8x)dx = 4 + C
b) ∫ 3 4 sin (x4 + 8x ) (x + 2)cos(x + 8x)dx = -----4-----+ C
c) ∫ 3 4 cos w (x + 2)cos(x + 8x)dx = 4 + C
d) This integral cannot be integrated using substitution.

 

Not correct. Choice (a) is false.
Try again, you have not integrated cosw correctly.
Your answer is correct.
Let w = x4 +8x then dw = (4x3 + 8)dx = 4(x3 + 2)dx and (x3 + 2)dx = 1dw . 4

∫ 3 4 (x + 2)cos(x + 8x)dx =∫ 1 1 4 cos w dw = 4 sin w
= 14sin(x4 + 8x)+ C
Not correct. Choice (c) is false.
Try again, you have not finished the integral.
Not correct. Choice (d) is false.
Try again, let w = x4 +8x .
 

Question 4

 
 
Which of the following are correctly evaluated?
a) ∫ 2 ∫ 9 1∕2 |9 x2(x3 + 1)1∕2 = w--- dw = 2w3 ∕2|| = 52 0 1 3 9 |1 9
b) ∫ π ∫ π |π θsin(θ2)dθ = w-dw = w2-|| = π2- 0 0 2 4 |0 4
c) ∫ 2 dx 1∫ -2 du - 1 ||-2 1 1 (3---5x)2-= - 5 -7 u2-= 5u-|| = 14 -7
d) ∫ 4√ ------ 1 ∫ 16√-- 1 2 3∕2||16 112 3x+ 4 dx = 3 w dw = 3 3w || = -9- 0 4 4
e) ∫ 1 ∫ -1 w w ||- 1 t2e-t3 dt = --e- dw = - e-|| = 1 - 1- 0 0 3 3 0 3 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.
2 ∫ π 2 ∫ π sinw-- 0 θsin(θ )dθ = 0 2 dw = |2 --cosw-||π 2 | 0
=1(cosπ2 - 1) 2
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.
∫ 2 dx 1∫ -7 du 1||-7 1 (3---5x)2-= - 5 u2-= 5u|| = 14 1 -2 -2
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.
Your answers are 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.
    2 ∫ π 2 ∫ π sinw-- 0 θsin(θ )dθ = 0 2 dw = |2 --cosw-||π 2 | 0
    =1(cosπ2 - 1) 2
  3. False. The answer is correct but the steps are incorrect. The limits of integration must stay in the correct order.
    ∫ 2 dx 1∫ -7 du 1||-7 1 (3---5x)2-= - 5 u2-= 5u|| = 14 1 -2 -2
  4. True.
  5. True.