Assignment 15

course Mth272

4/4/10 around 10:00am

015.

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Question: `qQuery problem 6.2.2 integrate x e^(-x)

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Your solution:

dv = e^(-x) v = e^(-x) u = x du = dx

uv – intg v – du

xe^(-x) – intg e^(-x)dx

x *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):

??????? Is it necessary to factor out the e^(-x). I understand the problem but did not factor out the e^(-x).

Generally algebraic expressions should be simplified, and in this case factoring out e^(-x) results in a simplified form.

However I usually don't penalize a correct answer for failure to simplify.

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Self-critique rating #$&*

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

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Your solution:

We let: dv = e^(-x) u = x^2 v = e^(-x) du = 2x*dx

u*v – intg v*du

x^2 * e^(-x) – intg 2x*e^(-x) 1st application by parts

then we let:

dv = e^(-x) u = 2x v = e^(-x) du = 2dx

x^2 * e^(-x) – 2x*e^(-x) – intg 2e^(-x) dx

x^2 *e(-x) – 2x*e^(-x) – 2e^(-x) + C

factor out –e^(-x)

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

confidence rating #$&*2

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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):

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Self-critique rating #$&*

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

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Your solution:

u = lnx du =1 / x

1 / u^3 = 1/ [u^(3+1) / 3 + 1 = 1 / (u^4 / 4)

1 / (u^4 / 4)

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):

??????????Having a hard time with the algebra on this one to get the answer

1/u^3 = u^(-3). The derivative of u^(-3) is -3 u^(-4).

However an antiderivative of u^(-3) is -1/2 u^-2.

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Self-critique rating #$&*

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

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Your solution:

dv = dx v = x u = ln(1+2x)

du = ln(1+2x)^(1+1) /1+1 = ln(1+2x)^2 / 2 = 2 ln (1=2x) / 2

confidence rating #$&*0

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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.

Self-critique (if necessary)

???????????I thought I had the right idea to us the power rule along with property 3 of the log rules to find du. After looking at your answer not sure where 2 / 1+2x comes from. Have a hard time again I guess with the algebra.

You have to use the chain rule, and the 'inner' function is (1 + 2 x), which has derivative 2.

More specifically:

ln(1 + 2x) is a composite of y = ln(z) and z = (1 + 2x). The derivative of z = (1 + 2x) with respect to x is just 2, so z ' = 2.

The derivative of the composite is (1 / z) * z ' = (1 / z) * 2 = 2 / z; since z = 1 + 2x we have derivative 2 / (1 + 2 z).

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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):

Having a hard time with algebra and seeing all the different rules to use in solving the problems.

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Self-critique rating #$&*

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