assignment 14

course Mth 272

july 22 10 pm

If your solution to a stated problem does not match the given solution, you should self-critique per instructions at

http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm.

Your solution, attempt at solution: If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.

014.

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Question: `qQuery problem 6.1.5 (was 6.1.4) integral of (2t-1)/(t^2-t+2)

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Your solution:

u = t^2 - t + 2. du = (2 t - 1) dt,

du / u.

=ln | u | + c

int( (2t-1) / (t^2 - 1 + 2) = ln t^2 - t + 2 + c.

Confidence rating: 2

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Given Solution: `a To integrate (2t-1) / (t^2-t+2) use u = t^2 - t + 2. We find that du = (2 t - 1) dt, which is there just waiting for you in the integrand.

This gives you integrand du / u.

The integral is ln | u | + c. Substituting we get

int( (2t-1) / (t^2 - 1 + 2) with respect to t) = ln | t^2 - t + 2 | + c.

The absolute value is important because t^2 - t + 2 can be negative.

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Self-critique (if necessary):

I didn’t realize I had to have the absolute value bars.

Self-critique Rating:

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Question: `qQuery problem 6.1.32 (was 6.1.26) integral of 1 / (`sqrt(x) + 1)

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Your solution:

u = (sqrt x) + 1----solve for x

x = (u-1)^2

dx = 2(u-1) du.-----dx=du

int(1 / (`sqrt(x) + 1) dx) the is int. (2 ( u - 1 ) / u) du

integrate ( u - 1) / u for u

( u - 1) / u = (u / u) - (1 / u) = 1 - (1 / u) u - ln | u |

substitute back in u = (sqrt x) + 1 and adding the 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 answer is 2* (2 (sqrt x) + 2 - 2 ln | (sqrt x) + 1 | + c.)

Final:

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

Confidence rating: 2

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Given Solution: `a 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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Self-critique (if necessary):

Self-critique Rating:

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Question: `qquery problem 6.1.56 (was 6.1.46) area bounded by x (1-x)^(1/3) and y = 0

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Your solution: Sketch graph to see the area bounded by the graph and by y = 0. Next, find the points where the graph crosses y = 0

x(1-x)^(1/3) = 0 when x = 0 or when 1-x = 0.

graph intersects at x = 0 and x = 1 (x axis)

integrate x (1-x)^(1/3) from 0 to 1. It becomes a definite integral.

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

integrand becomes (1 - u) * u^(1/3) = u^(1/3) - u^(4/3)

Anti-d= 3/4 u^(4/3) - 3/7 u^(7/3)= 3/4 (1-x)^(4/3) - 3/7 ( 1-x)^(7/3).

This=9/28 = about .32

Confidence rating: 2

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Given Solution: `a 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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Self-critique (if necessary):

Self-critique Rating:

Question: `qQuery problem P = int(1155/32 x^3(1-x)^(3/2), x, a, b).

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Your solution: int(1155/32 x^3(1-x)^(3/2), x, a, b)= int(1155 / 32 x^3 ( 1 - x)^(3/2), about x

Boundaries (limits) x = a and x = b

Confidence rating: 0

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Given Solution: `a For reference

int(1155/32 x^3(1-x)^(3/2), x, a, b) means 'the integral of 1155 / 32 x^3 ( 1 - x)^(3/2), integrated with respect to x

between limits x = a and x = b'.

That would be written with an integral sign with limits a and b, then 1155/32 x^3(1-x)^(3/2) dx.

It's easiest to integrate if you change the variable to u = 1 - x, which changes the integrand to 1155/32 (1 + u)^3 *

u^(3/2). Expand the cube and multiply through by u^(3/2) to get a sum of four power functions, easily integrated

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Self-critique (if necessary): I think I was on the right path, but got messed up at some point.

&#Your response did not agree with the given solution in all details, and you should therefore have addressed the discrepancy with a full self-critique, detailing the discrepancy and demonstrating exactly what you do and do not understand about the parts of the given solution on which your solution didn't agree, and if necessary asking specific questions (to which I will respond).

Self-critique Rating:

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Question: `qWhat is the probability that a sample will contain between 0% and 25% iron?

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Your solution: 50%

Confidence rating: 0

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Given Solution: `a 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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Self-critique (if necessary): Finding probability is one of the things I understand the least, if at all.

Self-critique Rating:

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Your solution: I cant answer this because I don’t know how to do probabilities. I have gone through them in the textbook, but no luck.

There was some confusion in the format of this query, which included a lot of followup questions which didn't have given solutions. This was also true of the subsequent queries. A little before midnight on 7/23/09 I edited the subsequent queries to eliminate these redundancies.

I've also edited your submission to eliminate these redundancies.

Can you tell me what you do and do not understand about probability distribution functions? If you do, then I can begin to help you understand them better.