assignment 15

course Mth 272

july 23 5 pm

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.

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

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Your solution:

u = x du = dx

dv = e^(-x)dx v = -e^(-x)

(x)(-e^(-x)) - int(-e^(-x)) dx….integrate that

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

pull out e^(-x):

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

Confidence rating: 2

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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.

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Self-critique (if necessary): Think I got this part pretty well.

Self-critique Rating:

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Question: `qWhat is the indefinite integral?

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Your solution: e^(-x)(-x-1)+C

Confidence rating: 1.5

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Given Solution

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Self-critique (if necessary):

Self-critique Rating:

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Question: `qWhat did you use for u and what for dv, what were du and v, and what integral did you obtain?

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Your solution:

u = x du = dx dv = e^(-x)dx v = -e^(-x)

Confidence rating:

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Given Solution:

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Self-critique (if necessary):

Self-critique Rating:

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

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Your solution:

X^2e^(-x)

two integrations by parts for two different parts.

u=x^2 dv=e^-x)dx v= -e^(-x)

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

int. x e^-x dx:

u=xd v=e^(-x)dx v= -e^(-x)

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

Substitute back -x^(2)e^(-x) +2int(xe^(-x) dx)

Answer: -x^(2)e^(-x) + 2 ( -x e^-x - e^-x + C) =-e^(-x) * (x^(2) + 2x +2)+ C.

Confidence rating: 2.5

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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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Self-critique (if necessary):

Self-critique Rating:

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Question: `qWhat is the indefinite integral?

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Your solution: -x^(2)e^(-x) + 2 ( -x e^-x - e^-x + C) =-e^(-x) * [x^(2) + 2x +2] + C

Confidence rating: 2-3

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Given Solution:

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Self-critique (if necessary):

Self-critique Rating:

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Question: `qFor your first step, what was u and what was dv, what were du and v, and what integral did you obtain?

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Your solution:

u=x

dv=e^(-x)dx

v= -e^(-x)

Confidence rating: 1

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution:

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Self-critique (if necessary):

Self-critique Rating:

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Question: `qAnswer the same questions for your second step.

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Your solution: -x e^(-x) - int(-e^(-x) dx) = -x e^(-x) - e^(-x) + c

Confidence rating: 1

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Given Solution:

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Self-critique (if necessary):

Self-critique Rating:

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

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Your solution:

1/(ln(x))^3)

u = ln(x)

du = 1 / x dx. =

1 / u^3 * du

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

Substituting u = ln(x) we have

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

Confidence rating: 2

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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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Self-critique (if necessary):

Self-critique Rating:

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Question: `qWhat is the indefinite integral?

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Your solution: -1 / (2 ln(x)^2) + c

Confidence rating:2

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Given Solution:

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Self-critique (if necessary):

Self-critique Rating:

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Question: `qWhat did you use for u and what for dv, what were du and v, and what integral did you obtain?

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Your solution: u=ln(x)

Confidence rating:

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Given Solution:

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Self-critique (if necessary):

Self-critique Rating:

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

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Your solution:

Confidence rating:

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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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Self-critique (if necessary):

&#You did not answer the given question. You need to always at least explain what you do and do not understand about the question. A phrase-by-phrase analysis is generally required when you cannot otherwise answer a question.

Self-critique Rating:

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Question: `qWhat is the indefinite integral?

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Your solution:

Confidence rating:

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Given Solution:

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Self-critique (if necessary):

Self-critique Rating:

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Question: `qWhat did you use for u and what for dv, what were du and v, and what integral did you obtain?

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Your solution:

Confidence rating:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution:

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Self-critique (if necessary):

Self-critique Rating:

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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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Self-critique (if necessary):

assignment 15

&#You did not answer the given question. You need to always at least explain what you do and do not understand about the question. A phrase-by-phrase analysis is generally required when you cannot otherwise answer a question.

&#Good work. See my notes and let me know if you have questions. &#