Assignment 14

course Mth 272

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19:55:57

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

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RESPONSE -->

confidence assessment:

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19:56:01

What is your result?

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integral of (2t-1)/(t^2-t+2) = ln|t^2-t+2|+C

confidence assessment: 2

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19:56:16

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

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Used the general log rule. u = t^2-t+2, du/dx = 2t-1

confidence assessment: 2

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19:56:20

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

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confidence assessment: 2

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19:56:33

What is your result?

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

confidence assessment: 2

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19:56:46

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

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RESPONSE -->

u = sqrt(x) + 1

x = (u-1)^2

dx/du = 2(u-1)

dx = 2(u-1)du

integral of 1 / (`sqrt(x) + 1) = 2(u-1)/u du = (2u-2)/u du = (2 - 2/u)du

= 2u - 2ln|u| + C

confidence assessment: 2

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19:56:58

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

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RESPONSE -->

du = dx/2(u-1)

x = (u-1)^2

confidence assessment: 2

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19:57:09

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 -->

dx = 2(u-1) du

confidence assessment: 2

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19:57:19

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 -->

2u - 2ln|u| + C

confidence assessment: 2

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19:57:29

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

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RESPONSE -->

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

confidence assessment: 2

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19:57:33

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

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confidence assessment:

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19:57:43

What is the area?

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A = .3214 square units

confidence assessment: 2

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19:57:53

What integral did you evaluate to obtain the area?

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-3 int. (1-u^3)(u^3)

-3 int. (u^3-u^6)

-3(1/4u^4 - 1/7 u^7)0,1

confidence assessment: 2

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19:58:05

What substitution did you use to evaluate the integral?

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u = (1-x)^1/3

x = 1-u^3

u^3 = 1-x

dx = -3u^2 du

confidence assessment: 2

Very good.

Just for comparison:

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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19:58:08

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

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confidence assessment:

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19:58:19

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

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43.27%

confidence assessment: 2

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19:58:29

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

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44.14%

confidence assessment: 2

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19:58:47

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

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RESPONSE -->

u=(1-x)^3/2

x = 1-u^2/3

dx = (-2/3) (u^-1/3) du

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

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

= 3465/64 int.(1-u^2/3)(-u^2/3) du

= 3465/64 int.(u^4/3-u^2/3)du

= 3465/64(3/7(u^7/3) - 3/2(u^2/3)

when x = 0, u = 1

x = .25, u = .6495

x = .5, u = .3536

x = 1, u = 0

when you plug the u values into the intergral you get .4324 and .4414.

confidence assessment: 2

I can't see an obvious flaw in your solution, but compare with the following:

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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19:58:50

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