Assignment 15

course Mth 272

I think I am close to getting lost. Please help with good comments and suggestions.

??}???????????assignment #015

015.

Applied Calculus II

10-18-2008

problem 6.2.2 integrate x e^(-x)

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17:52:43

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23:23:59

What is the indefinite integral?

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RESPONSE -->

xe^x-e^x+C

confidence assessment: 2

Good. For reference:

** 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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23:24:18

What did you use for u and what for dv, what were du and v, and what integral did you obtain?

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RESPONSE -->

u=x

dv= e^-x

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

du=dx

confidence assessment: 3

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23:24:37

Query problem 6.2.3 integrate x^2 e^(-x)

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RESPONSE -->

x^2e^-x - 2xe^-x + 2e^-x + C

confidence assessment: 1

Again good, though you missed a couple of signs.

** 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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23:24:55

What is the indefinite integral?

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RESPONSE -->

x^2e^-x= x^2e^-x - 2xe^-x + C

confidence assessment: 1

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23:25:15

For 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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RESPONSE -->

u= x^2

dv= e^-x dx

du= 2x dx

v= e^-x

If v = e^(-x) then dv = - e^(-x), which has the wrong sign.

v should be - e^(-x).

This error is what messed up the signs in your solution. Your solution was otherwise very good.

confidence assessment: 2

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23:25:37

Answer the same questions for your second step.

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RESPONSE -->

x^2e^-x - 2xe^-x

u= 2x

dv= e^-x dx

du= 2

v= e^-x

x^2e^-x - 2xe^-x + 2e^-x + C

confidence assessment: 2

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**** Query problem 6.2.18 integral of 1 / (x (ln(x))^3)

23:28:07

What is the indefinite integral?

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RESPONSE -->

(lnx^3x- 2x^2/x^3 + C)^-1

confidence assessment: 1

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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23:29:00

What did you use for u and what for dv, what were du and v, and what integral did you obtain?

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RESPONSE -->

u= lnx^3

dv= x

You don't have x; the x is in the denominator. Had you used dv = 1 / x you would have obtained v = ln(x), and with the simplified du (see below) it would have worked.

du= 1/x^3 * 2x^2 = 2x^2/x^3

This simplifies to 2 / x.

v= x

confidence assessment: 3

You weren't far wrong on this one; good approach, which with correct details would have worked.

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**** Query problem 6.2.34 integral of ln(1+2x)

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23:30:19

What is the indefinite integral?

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RESPONSE -->

ln(2x+1) dx= ln(2x+1)^2 - 2

= -ln(2x+1) dx + ln(2x+1)^2 - 2

confidence assessment: 1

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23:30:40

What did you use for u and what for dv, what were du and v, and what integral did you obtain?

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RESPONSE -->

u = ln(2x+1)

dv= dx

du= 2/(1+2x)

v= 2x+1

confidence assessment: 3

Good, but if dv = dx you end up with v = x, not v = 2x + 1.

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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23:33:52

Query Add comments on any surprises or insights you experienced as a result of this assignment.

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RESPONSE -->

I did not understand the format of this assignment. The material was very difficult, and no answers were provided after the submissions to check my work. I read through the chapter and examined the examples within the chapter before attempting this assignment, but some things just did not make sense. You can probably see that in my responses.

For the second question, my answer did not match the answer in the back of the book. I am not sure why. I thought I did that one right.

There is a lot I need help with on this. Please add as many comments as you can to my work. If it is not clear, I'd be glad to submit things in a more clear way somehow. Please let me know. Also, why aren't there any solutions in the assignment?

confidence assessment: 3

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You got the first problem, and you weren't far wrong on the second.

You didn't miss the third by much.

&#Let me know if you have questions. &#