Assignment 14

course Mth 272

7/28 12:30 pm

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: To integrate this expression, we need to label u and find du. Since the denominator of the expression has the highest power, (t^2-t+2) becomes u. When we take the integral of u, we find that it is equal to the numerator, 2t-1. So now we can rewrite our expression as du/u, making the integral equal to ln(abs(u)) + C. We can substitute back in for u, which makes our final integral ln(abs(t^2-t+2)) + C.

confidence rating: 3

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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):

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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: To solve the integral, we need to find u and du. We can let u equal the denominator of the fraction, so u = sqrt(x) +1. To make te expression simpler, we can then rewrite u to find x. We find that x equals (u-1)^2 and its derivative equals 2(u-1). We can now rewrite our whole integral to 2(u-1)/u, and when we integrate this according to our integration rules, we end up with 2(u-ln(abs(u))). Now we can plug back in our original value for u, ending up with the final integral of 2(sqrt(x) +1 – ln(abs(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): I had to work this problem a couple of ways before I finally got it right. I had trouble at the step of creating the new integral using only u as a variable.

This problem is a bit tricky. It's good that you understand.

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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: Since we are given the graph of this expression and we know that one of the bounds of the function is y = 0, we can easily identify that the domain of the function is [0,1]. Now we can integrate our function and then use the Fundamental Theorem of Calculus to find the area. Now we can integrate the expression x (1-x)^(1/3) according to our integration rules. First we can identify u as 1-x. Since the derivative of u is 1, we find that du = dx, and x = 1-u. Now we can rewrite our integral as (1 - u)(u^(1/3)) = (u^(1/3)) – (u^(4/3)). We can evaluate this using our integral rules, and we end up with 3/4 u^(4/3) - 3/7 u^(7/3). Now we can plug back in our value for u to get the final integral. Our final integral becomes 3/4 (1-x)^(4/3) - 3/7 ( 1-x)^(7/3). Finally, we can apply the Fundamental Theorem of Calculus to get the area. So (3/4 (1-1)^(4/3) - 3/7 ( 1-1)^(7/3)) – (3/4 (1-0)^(4/3) - 3/7 ( 1-0)^(7/3)) = 9/28.

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):

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Self-critique Rating:

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

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Your solution: To solve this integral, we first need to use u substitution. We find that the quantity of x-1 = u, making our new expression 1155/32 ((1 + u)^3)(u^(3/2)). We can now simply expand the expression using multiplication and take the integral of that easily. We find that the expanded expression is 1155/32(3+6u+3u^2+3u^(2/3)+3u^(4/3)). When we take the integral of that expression, we get 1155/32(3u+3u^2+u^3+9/5U^(5/3)+9/7u^(7/3)) +C.

confidence rating: 2

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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

(1 + u)^3 = 1 + 3 u + 3 u^2 + u^3, so

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

= 1155 / 32 ( u^(3/2) + 3 u^(5/2) + 3 u^(7/2) + u^(9/2)).

An antiderivative is

1155 / 32 ( 2/5 u^(5/2) + 6/7 u^(7/2) + 6/9 u^(9/2) + 2/11 u^(11/2) ).

Since u = 1 - x the limits on the u integral would be 1 - a and 1 - b.

The result would therefore be

1155 / 32 ( 2/5 (1-b)^(5/2) + 6/7 (1-b)^(7/2) + 6/9 (1-b)^(9/2) + 2/11 (1-b)^(11/2) ) - 1155 / 32 ( 2/5 (1-a)^(5/2) + 6/7 (1-a)^(7/2) + 6/9 (1-a)^(9/2) + 2/11 (1-a)^(11/2) ), which is slightly simplified to

1155 / 32 ( 2/5 (1-b)^(5/2) + 6/7 (1-b)^(7/2) + 6/9 (1-b)^(9/2) + 2/11 (1-b)^(11/2) ) - 2/5 (1-a)^(5/2) - 6/7 (1-a)^(7/2) - 6/9 (1-a)^(9/2) - 2/11 (1-a)^(11/2) ).

Factoring (1 - b)^(5/2) from the first four terms and (1 - a)^(5/2) from the last four terms, and simplifying yields

((1 - a)^(5/2)(105 a^3 + 70 a^2 + 40 a + 16) - (1 - b)^(5/2) (105 b^3 + 70 b^2 + 40 b + 16))/16.

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Self-critique (if necessary): Since this problem was so hairy when it was multiplied out, I’m not sure if I worked it out correctly. When I checked it against the given answer, -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) ), I found I was a little bit off.

That solution does get a bit messy, but I don't see anything obviously wrong with your result.

I've inserted the rest of the solution above.

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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: Since we know that our expanded integral is equal to -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) ), we can now simply solve for the probability. So -1155/32 * ( 2/5 ( 1 - .25 )^(5/2) - 3 * 2/7 ( 1 - .25 )^(7/2) + 3 * 2/9 * ( 1 - .25 )^(9/2) - 2/11 ( 1 - .25 )^(11/2) ) – (-1155/32 * ( 2/5 ( 1 - 0 )^(5/2) - 3 * 2/7 ( 1 - 0 )^(7/2) + 3 * 2/9 * ( 1 - 0 )^(9/2) - 2/11 ( 1 - 0 )^(11/2) )) = .0252 = 2.52%

confidence rating:

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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):

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Self-critique Rating:

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Question: `qWhat is the probability that a sample will contain between 50% and 100% iron?

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Your solution: Since we know that our expanded integral is equal to -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) ), we can now simply solve for the probability. So -1155/32 * ( 2/5 ( 1 - 1 )^(5/2) - 3 * 2/7 ( 1 - 1 )^(7/2) + 3 * 2/9 * ( 1 - 1 )^(9/2) - 2/11 ( 1 - 1 )^(11/2) ) – (-1155/32 * ( 2/5 ( 1 - .5 )^(5/2) - 3 * 2/7 ( 1 – .5)^(7/2) + 3 * 2/9 * ( 1 - .5 )^(9/2) - 2/11 ( 1 - .5 )^(11/2) ) = .736 = 73.6%.

confidence rating:

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Given Solution:

`a 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):

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