Due Friday January 17 by 3:00pm.

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Instructor (circle one): Robert Smith?

Frithjof Lutscher

DGD (circle one): 1

2

3

Student Name

Catalin Rada

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.

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