Assignment 15

course mth 272

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 6.1.6 (was 6.2.2) integrate x e^(-x)

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Your solution:

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

then factor out (e^(-x)

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

confidence rating #$&*

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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):

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Self-critique rating #$&*

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

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Your solution:

u=x^2

du = 2x dx

dv=e^-x dx

v= -e^(-x)

then

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

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

x e^-x dx:

u=x

dv = e^(-x)dx

v= -e^(-x)

Then the answer is

e^(-x)[– x ^(2) – 2x – 2 ] + C.

confidence rating #$&*3

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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):3

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Self-critique rating #$&*

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

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Your solution:

u = ln(x)

du = 1/x dx

sub back in and you get

1/u^3 du

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

.

confidence rating #$&*3

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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):

I understand this

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Self-critique rating #$&*3

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

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Your solution:

u = ln (1 + 2x)

du = 2/(2x+1) dx

(2x + 1)(ln(2x+1))/2 – x

Then plug in 0 and 1 then you get .648

confidence rating #$&*3

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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):

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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):

This was ok

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Self-critique rating #$&*

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&#Your work looks good. Let me know if you have any questions. &#