61--Assignment 14

course Mth 272

assignment #014

014.
Applied Calculus II
07-19-2008

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23:04:37
Query problem 6.1.5 (was 6.1.4) integral of (2t-1)/(t^2-t+2)

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RESPONSE -->
u=t^2-t+2
du/dt = 2t - 1
2t-1 ? = 2t-1
? = 1
(1) (1/u) du
ln (u) + C
ln(t^2-t+2) + C
confidence assessment: 2

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23:05:00
What is your result?

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RESPONSE -->
ln(t^2-t+2) + c
confidence assessment: 2

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23:05:35
What substitution did you use and what was the integral after substitution?

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RESPONSE -->
u = t^2-t+2
1/u
confidence assessment: 2

Good solution.

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23:11:26

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

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RESPONSE -->

u= x^(1/2)

du/dx = 1/ (2sqrt x)

(1/u) (2 sqrt x) du

2 sqrt x ln( sqrt x + 1)

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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23:12:14

What is your result?

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RESPONSE -->

2sqrt x [ ln (sqrt x + 1)]

confidence assessment: 2

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23:12:57

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

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RESPONSE -->

u= sqrt x +1

2sqrt x (1/u)

confidence assessment: 2

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23:14:25

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

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RESPONSE -->

(u+1)^2

confidence assessment: 1

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23:14:50

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 -->

2 sqrt x

confidence assessment: 2

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23:15:14

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 -->

2 sqrt x (1/u)

confidence assessment: 1

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23:15:54

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

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RESPONSE -->

2 sqrt x [ln (sqrt x) + 1]

confidence assessment: 1

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23:24:45

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 -->

u= 1 - x

x= 1 - u

x (u^(1/3))

(1 - u) (u^(1/3))

[u^(1/3)] - [u^(4/3)]

[(u^(4/3))/(4/3)] - [(u^(7/3))/(7/3)]

[(3/4)((1-x)^(4/3)] - [(3/7)((1-x)^(7/3)]

Evaluate from 0 to 1

= 0.32

confidence assessment: 2

Good, but be sure to document how you found the intersection points of the two curves; e.g.,

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.

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23:24:53

What is the area?

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RESPONSE -->

.32

confidence assessment: 2

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23:25:18

What integral did you evaluate to obtain the area?

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RESPONSE -->

[(3/4)((1-x)^(4/3)] - [(3/7)((1-x)^(7/3)]

Evaluate from 0 to 1

= 0.32

confidence assessment: 2

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23:25:37

What substitution did you use to evaluate the integral?

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RESPONSE -->

u= 1 - x

x= 1 - u

x (u^(1/3))

(1 - u) (u^(1/3))

[u^(1/3)] - [u^(4/3)]

[(u^(4/3))/(4/3)] - [(u^(7/3))/(7/3)]

confidence assessment: 2

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23:27:00

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

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RESPONSE -->

confidence assessment: 0

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23:28:17

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

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RESPONSE -->

need to go back to the question, but can't

confidence assessment: 0

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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23:28:25

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

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RESPONSE -->

need to go back to the question, but can't

confidence assessment: 2

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23:28:32

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

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RESPONSE -->

need to go back to the question, but can't

confidence assessment: 0

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23:31:37

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

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RESPONSE -->

1) Did not show me correct answers to allow me to check my work and give responses to the correct answers.

2) The last question waswritten with all kinds of commas and not very clear.

3)The assignment #'s are incorrect on the assignments sheet. This was assignment #14, but on the sheet it's listed as 15 and 16..I think this one covered 6.1 Part 1 AND 6.1 Part 2

Sorry for all the comments I just want to make sure I'm doing what's needed and try to do it correctly.

confidence assessment: 3

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You're doing fine. I'll check the assignment correspondences; email me if they aren't corrected in a day or two.

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