# Assignment 1 Solutions Essay

Submitted By calieh
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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

4

Student Number

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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.