asst 15

assignment #015

015.

Applied Calculus II

07-17-2008

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11:55:38

Query problem 6.2.2 integrate x e^(-x)

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RESPONSE -->

let dv = e^(-x)dx v = integral of dv = integral of e^(-x)dx = e^(-x)

u = x du =dx

integral of x e^(-x) = x e^(-x) - integral of e^(-x)

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

confidence assessment: 2

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11:56:55

What is the indefinite integral?

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RESPONSE -->

x e(-x) - Integral of e(-x) dx

confidence assessment: 2

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11:57:58

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 -->

these have been answered in previous answer

confidence assessment: 2

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12:08:23

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

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RESPONSE -->

dv = e^(-x) v= integral of dv = integral of e^(-x) dx = e^(-x).

u = x^2 du = 2x dx

x^2 e^(-x) =x^2 e^(-x) - integral of 2x e^(-x)

dv = e^(-x) v= integral of dv = integral of e^(-x) dx = e^(-x).

u = x^2 du = 2x dx

2nd application of integration by parts

x^2 e^(-x) =x^2 e^(-x) - integral of 2x e^(-x)

x^2 e^(-x) = x^2 e^(-x) - (2x e^(-x) - integral of 2 e^(-x) dx

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

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

confidence assessment: 2

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12:08:46

What is the indefinite integral?

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RESPONSE -->

answered in previous problem

confidence assessment: 2

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12:09:10

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 in previous answer

confidence assessment: 3

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12:09:36

Answer the same questions for your second step.

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RESPONSE -->

answered in previous answer

confidence assessment: 3

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12:22:27

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

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RESPONSE -->

= x(ln(x))^-3

dv = x^-3 v = integral of dv = integral of x^-3 dx = x^-2/-2

Good attempt, but x^-3 does not appear in this integrand.

u= ln x du= 1/x dx

integral of x(ln(x))^-3 = x^-2/-2 (ln x) - integral (x^-2/-2) (1/x) dx

= x^-2/-2 (ln x) - 1/2 integral of x dx

=x^-2/-2 (ln x) - x/2 + C

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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12:22:39

What is the indefinite integral?

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RESPONSE -->

already answered

confidence assessment: 3

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12:22:49

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 -->

alrready answered

confidence assessment: 3

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12:32:29

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

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RESPONSE -->

dv = dx dx=x

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

the derivative of ln(1 + 2x) is 2 / (1 + 2x) (chain rule)

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

good except for the factor of 2

= x ln(1+2x) - integral of 1/3dx

(x)(1/1+2x) is not 1/ 3 dx. Possible error in grouping?

=x ln(1+2x) - 1/3x + C

confidence assessment: 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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12:32:38

What is the indefinite integral?

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RESPONSE -->

already answered

confidence assessment: 2

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12:32:46

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 -->

already answered

confidence assessment: 2

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Good attempts; some errors in details. See my notes.