MAT 1332, Winter 2014, Assignment 1
Due Friday January 17 by 3:00pm.
Late assignments will not be accepted; nor will unstapled assignments.
Professors in the math department will not lend you a stapler; do not ask for one.
Instructor (circle one): Robert Smith?

Frithjof Lutscher

DGD (circle one): 1

2

3

Student Name

Catalin Rada
4

Student Number

By signing below, you declare that this work was your own and that you have not copied from any other individual or other source.
Signature

Question 1. Find the indefinite integral arcsin(x)dx. First Solution: Begin with integration by parts and use substitution afterwards.
For integration by parts, we set f (x) = arcsin(x) and g (x) = 1. Then f (x) = (1 − x2 )−1/2 and g(x) = x. Hence arcsin(x)dx = x arcsin(x) −

√

x dx. 1 − x2

Now we substitute u(x) = 1 − x2 so that du = −2xdx. Then we find
√
x du √ = u + C.
− √ dx =
2
2 u
1−x
Altogether, we obtain

arcsin(x)dx = x arcsin(x) +

1 − x2 + C.

Second Solution: We can also use substitution first and integration by parts afterwards.
Substitute y(x) = arcsin(x), or sin(y) = x. Typically, we differentiate the first form to find the differential, but here it is advantageous to differentiate the second and obtain cos(y)dy = dx.
Then the integral becomes arcsin(x)dx =

y cos(y)dy.

The latter integral is clearly solves by integration by parts with f (y) = y, g (y) = cos(y) so that y cos(y)dy = y sin(y) − sin(y)dy = y sin(y) + cos(y) + C.
Now we substitute back and find the same solution as above. Note cos(y) =

1 − sin2 (y) =

1

1 − x2 .

Question 2. Find the definite integral
2

3

x5 e−x dx.

0

First we substitute y = x3 . Then dy = 3x2 dx and x = 0 becomes y = 0, while x = 2 becomes y = 8. We get
2
8 y −y
3
x5 e−x dx = e dy
0
0 3
The latter integral is solved via integration by parts as
8
0

8

y y −y e dy = − e−y |80 +
3
3

0

1 −y y 1 e dy = − e−y |80 − e−y |80 .
3
3
3

Finally, we evaluate this expression as
2
0

1
1
3 x5 e−x dx = − (y + 1)e−y |80 = − 3e−8 .
3
3

Question 3. Consider the definite integral
2

ln(x)dx.

1

(a) Find the value of L4 , the Riemann sum approximation of the integral with four subintervals and function evaluation at the left hand endpoint of each subinterval.
L4 =

1
[ln(1) + ln(1.25) + ln(1.5) + ln(1.75)] ≈ 0.297
4

(b) Find the value of L10 .
L10 =

1
[ln(1) + ln(1.1) + ln(1.2) + ln(1.3) + · · · + ln(1.9)] ≈ 0.351
10

2

(c) Use the fundamental theorem to calculate the definite integral and compare its value with the two approximations.
Answer:
2
1

ln(x)dx = x ln(x)|21 −

2
1

dx = [x ln(x) − x]|21 = 2 ln(2) − 1 ≈ 0.386

Both Riemann sum approximations underestimate the