If your solution to a stated problem does not match the given solution, you should self-critique per instructions at  

http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm.

Your solution, attempt at solution: If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.

015.

Question: `qQuery problem 6.1.6 (was 6.2.2) integrate x e^(-x)

Your solution:

Confidence Assessment:

Given Solution:

`a We let

u = x

du = dx

dv = e^(-x)dx

v = -e^(-x)

Using u v - int(v du):

(x)(-e^(-x)) - int(-e^(-x)) dx

Integrate:

x(-e^(-x)) - (e^(-x)) + C

Factor out e^(-x):

e^(-x) (-x-1) + C.

Self-critique (if necessary):

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Question: `qQuery problem 6.1.7 (was 6.2.3) integrate x^2 e^(-x)

Your solution:

Confidence Assessment:

Given Solution:

`a We perform two integrations by parts.

First we use

u=x^2

dv=e^-x)dx

v= -e^(-x)

to obtain

-x^(2)e^(-x) - int [ -e^(-x) * 2x dx] =-x^(2)e^(-x) +2int[xe^(-x) dx]

We then integrate x e^-x dx:

u=x

dv=e^(-x)dx

v= -e^(-x)

from which we obtain

-x e^(-x) - int(-e^(-x) dx) = -x e^(-x) - e^(-x) + C

Substituting this back into

-x^(2)e^(-x) +2int[xe^(-x) dx] we obtain

-x^(2)e^(-x) + 2 ( -x e^-x - e^-x + C) =

-e^(-x) * [x^(2) + 2x +2] + C.

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Question: `qQuery problem 6.1.26 (was 6.2.18) integral of 1 / (x (ln(x))^3)

Your solution:

Confidence Assessment:

Given Solution:

`a Let u = ln(x) so that du = 1 / x dx. This gives you 1 / u^3 * du and the rest is straightforward:

1/u^3 is a power function so

int(1 / u^3 du) = -1 / (2 u^2) + c.

Substituting u = ln(x) we have

-1 / (2 ln(x)^2) + c.

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Question: `qQuery problem 6.1.46 (was 6.2.32) (was 6.2.34) integral of ln(1+2x)

Your solution:

Confidence Assessment:

Given Solution:

`a Let

u = ln ( 1 + 2x)

du = 2 / (1 + 2x) dx

dv = dx

v = x.

You get

u v - int(v du) = x ln(1+2x) - int( x * 2 / (1+2x) ) =

x ln(1+2x) - 2 int( x / (1+2x) ).

The integral is done by substituting w = 1 + 2x, so dw = 2 dx and dx = dw/2, and x = (w-1)/2.

Thus x / (1+2x) dx becomes { [ (w-1)/2 ] / w } dw/2 = { 1/4 - 1/(4w) } dw.

Antiderivative is w/4 - 1/4 ln(w), which becomes (2x) / 4 - 1/4 ln(1+2x).

So x ln(1+2x) - 2 int( x / (1+2x) ) becomes x ln(1+2x) - 2 [ (2x) / 4 - 1/4 ln(1+2x) ] or

x ln(1+2x) + ln(1+2x)/2 - x.

Integrating from x = 0 to x = 1 we obtain the result .648 approx.

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Question: `qQuery Add comments on any surprises or insights you experienced as a result of this assignment.

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