Assignment 15

course Mth 272

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Your work has been received. Please scroll through the document to see any inserted notes (inserted at the appropriate place in the document, in boldface) and a note at the end. The note at the end of the file will confirm that the file has been reviewed; be sure to read that note. If there is no note at the end, notify the instructor through the Submit Work form, and include the date of the posting to your access page.

20:04:02

Query problem 6.2.2 integrate x e^(-x)

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confidence assessment:

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20:04:17

What is the indefinite integral?

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

confidence assessment: 2

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20:04:38

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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u = x, du = dx

dv = e^-x dx, v = int. dv = int e^-x dx = -e^-x

confidence assessment: 2

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20:04:40

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

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confidence assessment:

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20:04:51

What is the indefinite integral?

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-x^2(e^-x) - 2x(e^-x) - 2(e^-x) + C

confidence assessment: 2

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20:05:02

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, du = 2x dx

dv = e^-x dx, v = intdv = int (e^-x dx) = -e^-x

x^2(e^-x)dx = x^2(-e^-x) - int.(-e^-x(2x dx)

= -x^2(e^-x) + int. 2x(e^-x) dx

confidence assessment: 2

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20:05:13

Answer the same questions for your second step.

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RESPONSE -->

u = 2x, du = 2

dv = e^-x, v = -e^-x

-x^2(e^-x) + int. 2x(e^-x) dx

-x^2(e^-x) - 2x(e^-x) - int.(e^-x)(2)

-x^2(e^-x) - 2x(e^-x) - 2(e^-x) + C

confidence assessment: 2

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20:05:17

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

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confidence assessment:

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20:05:27

What is the indefinite integral?

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ln(ln|x|)^3 + C

The derivative of this function is not 1 / (x (ln(x))^3).

confidence assessment: 2

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20:05:36

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|x|)^3, du = (ln|x|)^3

for this choice of u du would be 1/x * (2 ln|x|)^2

dv = (1/x) dx, v = ln x

confidence assessment: 2

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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20:05:39

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

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confidence assessment:

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20:05:49

What is the indefinite integral?

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RESPONSE -->

x ln(1+2x) - x + 2 ln(1+2x) + C

confidence assessment: 2

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20:06:00

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 = ln(1+2x), du = 1/(1+2x) dx

for this u you would get du = 2 / (1 + 2x) dx

dv = dx, v = x

integral of ln(1+2x) dx = ln(1+2x)dx - int. x(1/(1+2x))dx

= x ln(1+2x) - x + 2 ln(1+2x) + C

confidence assessment: 2

Very close, but the derivative of your result doesn't quite give you the original expression.

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.

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20:06:03

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

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confidence assessment:

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