School of Mathematics and Statistics
Junior
The University of Sydney
spcr

Quiz 4: Integration by Parts

Last unanswered question  Question  Next unanswered question
 

Question 1

 
 
Recall that integration by parts is a technique to re-express the integral of a product of two functions u  and ddvx  in a form which allows it to be more easily evaluated. The formula is ∫ u dv dx = uv - ∫ v dudx
    dx              dx  . When applying the method of integration by parts to find ∫ xex dx  , the best choice of u  and dv
dx  is
a) u = ex,  dv = x
        dx    b) u = xex,  dv= 1
          dx
c) u = 1, dv = xex
       dx    d) u = x,  dv-= ex
        dx
e) None of the above

 

Not correct. Choice (a) is false.
This expresses the original integral in terms of a new integral which is even harder! After applying integration by parts as suggested, we have ∫ xex dx = 1x2ex - ∫ 1x2ex dx
           2        2  .
Not correct. Choice (b) is false.
This expresses the original integral in terms of a new integral which is even harder! After applying integration by parts as suggested, we have ∫ xex dx = x2ex - ∫ (x2ex +xex) dx  .
Not correct. Choice (c) is false.
The problem of finding v  if dv = xex dx  is exactly the integral we started with! So no progress in this case.
Your answer is correct.
This gives genuine simplification and results in an easy integral. We obtain ∫ xex dx = xex - ∫ ex dx  .
Not correct. Choice (e) is false.
 

Question 2

 
 
Find ∫ 1xex dx
 0  using integration by parts, and enter your answer.

 

Your answer is correct
Choosing u = x,  dv= ex
        dx  gives ∫ 1xex dx = [xex]1- ∫1 ex dx = [ex(x- 1)]1= 1.
 0              0   0                 0
Not correct. You may try again.
Try choosing         dv   x
u = x,  dx = e  .
 

Question 3

 
 
The reduction formula for In = ∫ x(lnx)n dx  is In = 1x2(ln x)n - nIn-1
     2          2  . Given that I = 1x2 + C
 0  2  , find ∫e x(ln x) dx
 1  . Give your answer correct to three decimal places.

 

Your answer is correct
The integral I1  is 1 2      1 2
2x lnx - 4x + C  , using the reduction formula with n = 1  . Evaluating this between 1  and e  gives 2.097 to three decimal places.
Not correct. You may try again.
Substitute n = 1  into the reduction formula to obtain I1  .
 

Question 4

 
 
Which option is an antiderivative of x4ex   ? Use the reduction formula I  = ∫ xnex dx = xnex - nI
 n                      n-1  to help answer this question.
a) exx4 + 4x3ex - 12x2ex +24xex - 24ex    b) ex(x4 - 4x3 - 12x2 - 24x + 24)
c) exx4 - 4x3ex +12x2ex - 24xex + 24ex + Cex    d) ex(x4 - 4x3 + 12x2 - 24x+ 24)+ 15
e) None of the above

 

Not correct. Choice (a) is false.
Not correct. Choice (b) is false.
Not correct. Choice (c) is false.
Your answer is correct.
Not correct. Choice (e) is false.
 

Question 5

 
 
Which option equals ∫1 xtan-1x dx
 0   ? (Hint: use integration by parts with       -1
u = tan x  and dv
dx = x  .)
a) π-  1
4 - 2    b)  1
-2
c) π-  1
 8 - 2    d) 1 π-
2 (4 + 1)
e) π-
4

 

Your answer is correct.
The indefinite integral is 12x2 tan- 1x- 12x + 12 tan-1 x+ C  , which, when evaluated between 0 and 1, matches this option.
Not correct. Choice (b) is false.
Not correct. Choice (c) is false.
Not correct. Choice (d) is false.
Not correct. Choice (e) is false.
 

Question 6

 
 
Which option equals ∫ 12 sin- 1x dx
 0   ? (Hint: use integration by parts with u = sin-1 x  and ddvx = 1  .)
a)      √-
-π - -3-+ 1
12    2    b)     √-
π-+ -3-
4   2
c)      √-
-π + -3-- 1
12    2    d) π-- 1
3
e)      √ -
-π - --3- 1
12    2

 

Not correct. Choice (a) is false.
Not correct. Choice (b) is false.
Your answer is correct.
The indefinite integral is     -1    √-----2
x sin  x +  1 - x + C  , which, when evaluated between 0  and 12  , matches this option.
Not correct. Choice (d) is false.
Not correct. Choice (e) is false.
 

Question 7

 
 
The finite area bounded by the curve y = ln x  , the line y = 1  and the tangent line to y = ln x  at x = 1  is given as an integral with respect to x  by ∫                  ∫
 12(x- 1- lnx) dx+  e2 (1 - ln x) dx  . Which option equals the same area given as an integral with respect to y   ? (You must draw a sketch to help you with this question.)
a) ∫
 e1 ey - (y+ 1) dy    b) ∫
 10 ln y- y+ 1 dy
c) ∫
 0eey - y + 1 dy    d) ∫ 1ey - (y +1) dy
 0
e) None of the above

 

Not correct. Choice (a) is false.
Not correct. Choice (b) is false.
Not correct. Choice (c) is false.
Your answer is correct.
Not correct. Choice (e) is false.
 

Question 8

 
 
Which option equals ∫ xsec2x dx   ?
a) xsecxtan x- ln(cosx)+ C    b) x tan x+ ln∣cosx∣+ C
c) xtan2x - ln∣cosx∣+ C    d) xtan x- ln(cosx)+ C
e) None of the above

 

Not correct. Choice (a) is false.
Your answer is correct.
Not correct. Choice (c) is false.
Not correct. Choice (d) is false.
Not correct. Choice (e) is false.
 

Question 9

 
 
In some problems you need to apply the integration by parts method twice in order to obtain the required answer. The integral ∫
  sin(lnx) dx  is one such problem. Which of the following options gives the expression obtained after one application of integration by parts?
a)            ∫
x sin(ln x)-   cos(ln x) dx    b) - cosxlnx + ∫ cosx dx
               x
c)             ∫
- x sin(ln x)+  x cos(lnx) dx    d)            ∫
xsin(ln x)-   sin(ln-x)dx
                x
e)           ∫ cosx
cosx lnx +   --x- dx

 

Your answer is correct.
Not correct. Choice (b) is false.
Try u = sin(ln x)  and dv=  1
dx  in the integration by parts formula.
Not correct. Choice (c) is false.
Not correct. Choice (d) is false.
Not correct. Choice (e) is false.
 

Question 10

 
 
Which option equals ∫e
 1 sin(ln x) dx   ?
a) 1
2(esin 1+ ecos1)    b) sin1- cos1
c) 1
2 (e sin1 - ecos1+ 1)    d) 1
2 sin 1- 2ecos1+ 1
e) esin1 - ecos1

 

Not correct. Choice (a) is false.
Not correct. Choice (b) is false.
Your answer is correct.
Not correct. Choice (d) is false.
Not correct. Choice (e) is false.