Assignment 15

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course MTH272

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:

u=x

du=1

v=e^(-x)

dv=-e^(-x)dx

[x*-e^(-x)]-[int1*-e^-x dx]

-x-1(e^-x)( e^-x)+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):I came out with a negative x and I don’t understand what happened to it (x) in your end solution.

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Self-critique Rating:

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The derivative of e^(-x) ( -x - 1 ) is

(e^-x) ' ( -x - 1) + e^-x ( -x - 1) ' =

-e^(-x) ( -x - 1) - e^(-x) =

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

x e^(-x).

So the given solution is correct. I had to check it to be sure of all the signs; with more than a couple negative signs it's easy to make a mistake.

Your expression

[x*-e^(-x)]-[int1*-e^-x dx]

yields

-x e^(-x) + int(1 * e^(-x) dx)

which is equal to

-x e^(-x) - e^(-x)

and thus to

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

It appears you lost track of a negative sign in your solution. Otherwise it was fine.

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The derivative of e^(-x) ( -x - 1 ) is

(e^-x) ' ( -x - 1) + e^-x ( -x - 1) ' =

-e^(-x) ( -x - 1) - e^(-x) =

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

x e^(-x).

So the given solution is correct. I had to check it to be sure of all the signs; with more than a couple negative signs it's easy to make a mistake.

Your expression

[x*-e^(-x)]-[int1*-e^-x dx]

yields

-x e^(-x) + int(1 * e^(-x) dx)

which is equal to

-x e^(-x) - e^(-x)

and thus to

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

It appears you lost track of a negative sign in your solution. Otherwise it was fine.

*@

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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:

x^2 e^(-x)

u=x^2

du=2x

v=-e^-x

dv=e^-x

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

Repeated 2 int xe^(-x) dx

Then the second part

u=x

dv=e^(-x)dx

v= -e^(-x)

[-x e^(-x)] - [int(-e^(-x) dx)=-x e^(-x) - e^(-x) + c]

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

-x^(2)e^(-x)+[2(-x e^-x-e^-x+c)]

confidence rating #$&*:1

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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): if there had not been an example of this in the book I don’t know if I would have completed this all the way through.

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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:

1 / (x (ln(x))^3)

u=ln(x)

du=1/x

dx=(x)du

1/(u)^3=1/ln(x)^3

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

confidence rating #$&*:1

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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):NA

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Self-critique Rating:

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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:

ln(1+2x)

u= ln(1+2x)

du=2/1+2x (dx)

v=x

dv=dx

(x*ln(1+2x))-(int x*2/1+2x (dx))

(x*ln(1+2x))-(2int x/1+2x (dx))

(x*ln(1+2x))-[2(2x/4-1/4)(ln(1+2x))]

(x*ln(1+2x))+[ln(1+2x)/2(-x)]

x=0 x=1

[0*ln(1+2(0)]+[ln(1+2(0))/2(-0)]=0

[(1)*ln(1+2(1))]+[ln(1+2(1))/2(-1)]=.6479

.648 square units

confidence rating #$&*:1

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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):I had to dig in every direction on this one and it took some information from the web to get me through it.

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Self-critique Rating:

@&

You only have one term in the integrand, so you have only one option on this problem and that is to let u = ln(1 + 2x), dv = dx.

Having done that it isn't difficult to go through the steps.

However I question whether the text has introduced integration by parts in Section 6.1.

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

This was probably the toughest assignment yet for me

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Self-critique (if necessary):

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Self-critique Rating:

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Integration technique is, as I believe I mentioned in another note, pretty much a bag of tricks. That makes it challenging. However this course makes use of a fairly manageable number of tricks, the most basic of which can be learned by sufficient practice.

It's the trickier tricks that are giving you problems. That happens to everyone. You're doing fine with the basic ones. It does take awhile for the process to sink in.

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