Assignment 15

course Mth 272

?????Sw???????E??assignment #015015.

Applied Calculus II

03-13-2009

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20:52:50

Query problem 6.2.2 integrate x e^(-x)

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RESPONSE -->

first find all the parts

u=x

du=dx

dv=e^-x

v= -e^-x

then use the integ by parts formula

-xe^-x - e^-x +c then factor

(e^-x)(-x-1) +c

confidence assessment: 3

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20:53:01

What is the indefinite integral?

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RESPONSE -->

answered

confidence assessment: 3

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20:53:14

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 -->

answered

confidence assessment: 3

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21:08:45

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

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RESPONSE -->

finding the parts

u=x^2

du=2xdx

dv=e^-xdx

v=-e^-x

-x^2e^-x+2 ~ xe^-xdx

the parts are the same as the last question so i don't have to do them all over again

-x^2e^-x+2[-xe^-x-e^-x]+c

-x^2e^-x-2xe^-x-2e^-x+c factor

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

confidence assessment: 3

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21:08:53

What is the indefinite integral?

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RESPONSE -->

answered

confidence assessment: 3

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21:09:01

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 -->

answered

confidence assessment: 3

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21:09:11

Answer the same questions for your second step.

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RESPONSE -->

answered

confidence assessment: 3

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21:23:38

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

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RESPONSE -->

you actually don't have to use integration by parts for this one.

u= ln(x)

du= 1/x dx

~ 1/(lnx)^3dx = ~ lnx^-3 dx then use the power rule

lnx^-2/ -2

= -1/2lnx^2 + c

confidence assessment: 3

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21:23:47

What is the indefinite integral?

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RESPONSE -->

answered

confidence assessment: 3

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21:23:56

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 -->

answered

confidence assessment: 3

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21:37:41

Query problem 6.2.32 (was 6.2.34) integral of ln(1+2x)

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RESPONSE -->

this on doesn't require integration by parts either

u= 1+2x

du=2dx

1/2 ~ ln (1+2x)dx

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

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

i used the logarithmic rule that states the integral of lnxdx = xlnx- x + c

Your solution would be legitimate if you first derived the result you quote, lnxdx = xlnx- x + c.

Without using that result:

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.

confidence assessment: 3

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21:37:48

What is the indefinite integral?

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RESPONSE -->

answered

confidence assessment: 3

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21:37:55

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 -->

answered

confidence assessment: 3

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21:38:29

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

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RESPONSE -->

I understood all of this assignment, and it was helpful to see all the problems in the book.

confidence assessment: 3

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