hw asst 14

course mth 272

C????????????assignment #014

014.

Applied Calculus II

07-17-2008

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10:15:09

Query problem 6.1.5 (was 6.1.4) integral of (2t-1)/(t^2-t+2)

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RESPONSE -->

start by substituting u=(t^2-t+2) du/dt = 2t -1

not sure if you solve for t, or if you now go ahead and substitute for t and dt.

confidence assessment: 0

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10:15:36

What is your result?

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RESPONSE -->

confidence assessment: 0

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10:17:24

What substitution did you use and what was the integral after substitution?

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RESPONSE -->

i think.... u = (t^2 -t +2) du=2t -1

integral of 2t-1 / u

confidence assessment: 1

start by substituting u=(t^2-t+2) du/dt = 2t -1.

So du = (2t - 1) dt.

Thus your integrand (2t - 1) / (t^2 - t + 2) dt is equal to du / u.

Antiderivative is ln | u | + c = ln | t^2 - t + 2 | + c.

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10:27:53

Query problem 6.1.32 (was 6.1.26) integral of 1 / (`sqrt(x) + 1)

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RESPONSE -->

well, i was hoping to see how the last problem was worked and try and work backwards with it.

You were on the right track. Not sure what happened to that solution but I completed yours. See above.

next problem.

u ='sqrt(x) + 1 du= x^1/2 =1/2x^3/2

= integral of (1/u)du

=ln{u} + C

=ln('sqrt(x) + 1) + C

confidence assessment: 1

If we let u = (sqrt x) + 1 we can solve for x to get

x = (u-1)^2 so that

dx = 2(u-1) du.

So the integral of 1 / (`sqrt(x) + 1) dx becomes the integral of (2 ( u - 1 ) / u) du.

Integrating ( u - 1) / u with respect to u we express this as

( u - 1) / u = (u / u) - (1 / u) = 1 - (1 / u). An antiderivative is u - ln | u |.

Substituting u = (sqrt x) + 1 and adding the integration constant c we end up with

(sqrt x) + 1 - ln | (sqrt x) + 1 | + c.

Our integral is of 2 (u-1)/u, double the expression we just integrated, so our result will also be double. We get

2 (sqrt x) + 2 - 2 ln | (sqrt x) + 1 | + c.

Since c is an arbitrary constant, 2 + c is also an arbitrary constant so the final solution can be expressed as

2 (sqrt x) - 2 ln | (sqrt x) + 1 | + c.

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10:28:34

What is your result?

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RESPONSE -->

ln('sqrt(x) + 1) + C

confidence assessment: 2

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10:29:54

What substitution did you use and what was the integral after substitution?

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RESPONSE -->

i used the general log rule form of substitution.

the integral after substitution was written previously

confidence assessment: 2

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10:33:05

If you let u = `sqrt(x) + 1 then what is du and, in terms of u, what is x?

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RESPONSE -->

i think du= (x)^1/2 = 1/2 x^3/2

x = 2/3 *( u/1/2)

confidence assessment: 1

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10:35:22

If you solve du = 1 / (2 `sqrt(x)) dx for dx, then substitute your expression for x in terms of u, what do you finally get for dx?

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RESPONSE -->

i have no idea

confidence assessment: 0

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10:36:07

What therefore will be your integrand in terms of u and what will be your result, in terms of u?

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RESPONSE -->

confidence assessment: 0

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10:36:56

What do you get when you substitute `sqrt(x) + 1 for u into this final expression?

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RESPONSE -->

= ln('sqrt(x) + 1) + C

confidence assessment: 1

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10:50:27

query problem 6.1.56 (was 6.1.46) area bounded by x (1-x)^(1/3) and y = 0

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RESPONSE -->

why is there no critique in here for me to review.

u = 1 - x du = dx

x = 1 - u

integral of (1-u) (u^1/3) du

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

=(u^4/3 / 4/3) + (u^7/3 / 7/3) + C

= 3/4 (1-x)^4/3 + 7/3(1-x)^7/3 + C

confidence assessment: 2

Good, but the reciprocal of 7/3 is 3/7.

Begin by sketching a graph to see that there is a finite region bounded by the graph and by y = 0. Then find the points where the graph crosses the line y = 0 but finding where the expression takes the value 0:

x(1-x)^(1/3) = 0 when x = 0 or when 1-x = 0. Thus the graph intersects the x axis at x = 0 and x = 1.

We therefore integrate x (1-x)^(1/3) from 0 to 1.

We let u = 1-x so du = dx, and x = 1 - u.

This transforms the integrand to (1 - u) * u^(1/3) = u^(1/3) - u^(4/3), which can be integrated term by term, with each term being a power function.

Our antiderivative is 3/4 u^(4/3) - 3/7 u^(7/3), which translates to 3/4 (1-x)^(4/3) - 3/7 ( 1-x)^(7/3).

The result is 9/28 = .321 approx..

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10:54:15

What is the area?

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RESPONSE -->

use fundamental theorem for 0 and 1

3/4(1-1)^4/3 + 7/3(1-1)^7/3 + C =0

3/4(1-0)^4/3 + 7/3(1-0)^7/3 + C = 1.027

0-1.027

=-1.027

confidence assessment: 1

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10:55:04

What integral did you evaluate to obtain the area?

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RESPONSE -->

integral of (1-u)(u^1/3) du

confidence assessment: 2

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10:55:34

What substitution did you use to evaluate the integral?

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RESPONSE -->

let u = 1-x du =dx x=1-u

confidence assessment: 1

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11:08:41

Query problem P = int(1155/32 x^3(1-x)^(3/2), x, a, b).

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RESPONSE -->

u = (1-x)^3/2

u^3/2 = 1-x

x = 1-u^3/2 dx= -3/2 u^1/2

when x =0 u = (1-0)^3/2 =1

when x= .25 u = (1-.25)^3/2 = .211

integral at .211 and 1 of [ (1155/32) (1- u^3/2)(u)(-3/2u^1/2) ] du

305/16 integral at .211 and 1 of ( here I'm not sure)

confidence assessment: 1

The probability of an occurrence between 0 and .25 is found by integrating the expression from x = 0 to x = .25:

Let u = 1-x so du = -dx and x = 1-u.

Express in terms of u:

-(1155/32) * int ( (1-u)^3 (u)^(3/2) du )

Expand the integrand:

-(1155/32) * int( (1 - 3 u + 3 u^2 - u^3) * u^(3/2) ) =

-1155/32 * int( u^(3/2) - 3 u^(5/2) + 3 u^(7/2) - u^(9/2) ) .

An antiderivative of u^(3/2) - 3 u^(5/2) + 3 u^(7/2) - u^(9/2) is 2/5 u^(5/2) - 3 * 2/7 u^(7/2) + 3 * 2/9 * u^(9/2) - 2/11 u^(11/2) . So we obtain for the indefinite integral

-1155/32 * ( 2/5 u^(5/2) - 3 * 2/7 u^(7/2) + 3 * 2/9 * u^(9/2) - 2/11 u^(11/2) ).

Express in terms of x:

-1155/32 * ( 2/5 ( 1 - x )^(5/2) - 3 * 2/7 ( 1 - x )^(7/2) + 3 * 2/9 * ( 1 - x )^(9/2) - 2/11 ( 1 - x )^(11/2) )

Evaluate this antiderivative at the limits of integration 0 and .25 .

You get a probability of .0252, approx, which is about 2.52%, for the integral from 0 to .25; this is the probability of a result between 0 and 25%.

To get the probability of a result between 50% and 100% integrate between .50 and 1. You get .736, which is 73.6%.

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11:17:34

What is the probability that a sample will contain between 0% and 25% iron?

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RESPONSE -->

using integral from previous where I had already started for 0 and 25%...

(305/ 16) integral at .211 and 1 ( u^3/4 -u^3/2) du

305/16 [ (u^3/4 / 3/4) - (u^3/2 / 3/2) ] at .211 and 1

= -1.2279 for .211

= -.794 for 1

=-.4339

confidence assessment: 2

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11:22:21

What is the probability that a sample will contain between 50% and 100% iron?

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RESPONSE -->

when x =50 u = (1-50)^3/2 =343

when x =100 u = (1-100)^3/2 = 985.04

integral at 343 lower and 985 upper

then use same formula and apply fundamental theorem to obtain answer

confidence assessment: 2

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11:23:38

What substitution or substitutions did you use to perform the integration?

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RESPONSE -->

u =(1-x)^3/2

u^3/2 = 1 -x

x=1-u^3/2 dx= -3/2u^1/2

confidence assessment: 2

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11:25:04

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

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RESPONSE -->

i found this assignment to be the hardest assignment that I have had. In going through the examples in the book I understood everyone. But the assigned problems really tripped me up. it was pretty time consuming as well.

confidence assessment: 3

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