Assignment 14

course MTH 272

......!!!!!!!!...................................Applied Calculus II

Your work has been received. Please scroll through the document to see any inserted notes (inserted at the appropriate place in the document, in boldface) and a note at the end. The note at the end of the file will confirm that the file has been reviewed; be sure to read that note. If there is no note at the end, notify the instructor through the Submit Work form, and include the date of the posting to your access page.

Asst # 14

06-30-2006

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15:02:54

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

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15:05:20

u= t^2 -t +2

du = 2t -1

int ((du/dt)/u)

= ln |u| +C

= ln |t^2 -t +2| +C

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15:05:22

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**** What is your result?

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15:05:44

ln |t^2 -t +2| +C

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15:05:45

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

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15:06:28

u= t^2 -t +2

du = 2t -1

Integral after substitution: int ((du/dt)/u)

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15:06:29

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**** Query problem 6.1.13 (was 6.1.32) integral of 1 / (`sqrt(x) + 1)

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15:07:42

u = x + 1

du = 1

int u^-1/2

= 2u^1/2 +C

= 2 sqrt(u) + C

= 2 sqrt(x+1) +C

Your solution would be correct if the expression was 1 / (sqrt(x+1)). However 1 / (`sqrt(x) + 1) is a different expression; x + 1 is not part of this expression.

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15:07:43

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**** What is your result?

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15:07:52

= 2 sqrt(x+1) +C

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15:07:52

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

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15:08:05

u = x + 1

du = 1

Integral after substitution:

int u^-1/2

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15:08:06

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

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15:10:22

u = sqrt(x) +1

du= .5 x^(-.5)

x^.5 = u - 1

x = (u-1)^2

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15:10:22

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**** 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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15:14:58

I'm not sure if I understand the question correctly. If you solve du = 1 / (2 `sqrt(x)) dx for dx, then you get dx = 2 du / sqrt(x)

Substituting (u-1)^2 in for x, you get:

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

= 2 du / (u-1) = dx

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15:14:58

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

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15:20:08

int ((2 du) / (u-1))

= 2 int ((u-1)^(-1))

= 2 ln |u -1|

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

= 2 ln | sqrt x|

I don't know whether or not this answer is correct. This query program is showing up as very different than the program for assignments 1-13. It's not giving me the correct solutions after I submit my answers, so I have no clue if my answers are right or wrong, and I'd like to know the correct way to solve these problems so I won't make the same mistakes again in future problems.

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15:20:09

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

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15:20:46

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

= 2ln | sqrtx | + C

&& 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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15:20:47

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

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15:26:41

u = 1-x

du = -1dx

x = (1-u)

- int [0,1] (1-u)u^(1/3)

-int (u^1/3 - u^4/3)

-[3/4 u^(4/3) - (3/7)u^(7/3)) + C [0,1]

- [3/4 - 3/7)

- 9 /28

I think that this answer is wrong because clearly the area is not negative. Maybe it should just be 9/28.

When you change the variable to 1 - u, the original limits x = 0 and x = 1 become u = 1 and u = 0. So your integral would be from u = 1 to u = 0.

This would give you a positive integral, which would correspond to the fact that on the interval 0 < x < 1 the function lies entirely above the x axis.

The integral therefore does come out positive, and other than the limits and the resulting negative value, your work is entirely correct; 9/28 is the correct answer.

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15:26:43

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What is the area?

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15:27:19
I got - 9/28, but it may be + 9/28, or 9/28 may not be right at all, regardless of it being positive or negative.

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15:27:20

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What integral did you evaluate to obtain the area?

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15:27:48
the integral of u^(1/3) - u^(4/3)

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15:27:49

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What substitution did you use to evaluate the integral?

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15:28:05
u = 1-x
du = -1 dx
x = 1-u

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15:28:05

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Query problem 6.1.64 P = int(1155/32 x^3(1-x)^(3/2), x, a, b).

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15:40:53
a) between 0 and 25%
[0, .25] int ( 1155/32 x^3 (1-x)^(3/2)
u = 1-x
du = -1
x = 1-u
x^3 = (1-u)^3 = (-u^3 + 3u^2 - 3u +1)
-1155/32 int ((-u^3 + 3U^2 - 3u +1) u^(3/2))
-1155/32 int ((-u^(9/2) + 3u^(7/2) - 3u^(5/2) + u^(3/2)))
-1155/32 [ (-2/11u^(11/2) + (2/3)u^(9/2) - (6/7)u^(7/2) + (2/5)u^(5/2)]
-1155/32 (-(2/11)(1-x)^(11/2) + (2/3)(1-x)^(9/2) - (6/7)(1-x)^(7/2) + (2/5)(1-x)^(5/2)]
-1155/32 [(-.037 + .182677 - .31316 + .1949) - (-2/11 + 2/3 - 6/7 + 2/5)]
-1155/32 (.027417 - (32/1155))
-.98958 + 1
= .0104176563
= 1.04%
b) between 50% and 100%
-1155/32 (-(-.004 + .02946 - .07576 + .0707))
-1155/32 (-(.0204))
.7363125
= 73.6%

This is correct.

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15:40:54

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

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15:41:06

a) between 0 and 25%

[0, .25] int ( 1155/32 x^3 (1-x)^(3/2)

u = 1-x

du = -1

x = 1-u

x^3 = (1-u)^3 = (-u^3 + 3u^2 - 3u +1)

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

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

-1155/32 [ (-2/11u^(11/2) + (2/3)u^(9/2) - (6/7)u^(7/2) + (2/5)u^(5/2)]

-1155/32 (-(2/11)(1-x)^(11/2) + (2/3)(1-x)^(9/2) - (6/7)(1-x)^(7/2) + (2/5)(1-x)^(5/2)]

-1155/32 [(-.037 + .182677 - .31316 + .1949) - (-2/11 + 2/3 - 6/7 + 2/5)]

-1155/32 (.027417 - (32/1155))

-.98958 + 1

= .0104176563

= 1.04%

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15:41:06

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

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15:42:03

between 50% and 100%

u = 1-x

du = -1

x = 1-u

x^3 = (1-u)^3 = (-u^3 + 3u^2 - 3u +1)

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

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

-1155/32 [ (-2/11u^(11/2) + (2/3)u^(9/2) - (6/7)u^(7/2) + (2/5)u^(5/2)]

-1155/32 (-(2/11)(1-x)^(11/2) + (2/3)(1-x)^(9/2) - (6/7)(1-x)^(7/2) + (2/5)(1-x)^(5/2)]

-1155/32 (-(-.004 + .02946 - .07576 + .0707))

-1155/32 (-(.0204))

.7363125

= 73.6%

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15:42:04

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

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15:42:16

u = 1-x

du = -1

x = 1-u

x^3 = (1-u)^3 = (-u^3 + 3u^2 - 3u +1)

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15:42:16

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

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15:43:58

I'm still getting used to the new layout of the query program. I don't like not seeing the correct solutions, because I have a feeling many of my answers were incorrect. I can do some of the easier problems, but once they start getting more difficult, I have a much harder time getting the correct answer. I guess I just need more practice. I can understand the examples given in the book, but once I have problems I have to do on my own, lots of times, I don't even know where to begin.

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15:44:05

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15:44:19

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Very good work; do be sure to see my notes correcting the two errors you did make. Let me know if you have questions.

I will check the problem correspondences with the latest edition of the text; it is possible that some of the problem assignments refer to problem numbers from the previous edition.

From this point on, the queries do not include the solutions. However, as a result I will be inserting many more comments into your submitted work.