Assignment 14

course Mth 272

?y???????uF??assignment #014

014.

Applied Calculus II

03-13-2009

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18:41:50

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

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RESPONSE -->

using u substitution, u= T^2-t+2 and du- 2t-1 that in an indication that the general log rule should be used because by rewriting it we get du/u which = 1/u* du

so

ln[u] + c then replace the terms, so the integral is

ln[t^2-t+2]+c

confidence assessment: 3

Good, but it's ln | t^2-t+2 | + c. The absolute value is important because t^2 - t + 2 can be negative.

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18:49:11

What is your result?

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RESPONSE -->

ln[t^2-t+2]+c

confidence assessment: 3

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18:50:32

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

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RESPONSE -->

i used normal u substitution. i didn't have to figure out dx or x specifically, because the problem was very simple. i used the general log rule also

confidence assessment: 3

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19:02:42

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

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RESPONSE -->

first find u = sqrt(x) + 1

u--1= sqrt(x)

so x= (u-1)^2

dx= 2sqrt(x)du by multiplying

so in terms of u its

1/ sqrt(u-1)^2 + 1 du replace the u's with terms and use general log rule

= 2sqrt(x) - 2ln[sqrt(x) +1] + c

confidence assessment: 3

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19:02:47

What is your result?

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RESPONSE -->

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

confidence assessment: 3

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19:03:36

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

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RESPONSE -->

i used integration by substitution, and the general log rule again

shown in last problem

confidence assessment: 3

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19:05:55

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 = 1/2sqrt(x) dx

x= (u-1)^2

dx= 2sqrt(x)du

confidence assessment: 3

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19:08:18

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 -->

1/sqrt(u-1)^2 + 1 du

dx= 2sqrt(x)du

confidence assessment: 3

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19:08:32

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 -->

answered previously

confidence assessment: 3

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19:08:56

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

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RESPONSE -->

previously answered

confidence assessment: 3

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19:55:27

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 -->

y= xcubert(1-x) dx y= [0,1]

interval is from 0 to 1 x(1-x)^1/3dx is rewritten

u= 1-x

du= -dx

x- 1-u

since its negative you get new limits u= 1-0 = 1 because when x=0 u = 1 you can change it to positive

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

+{ from 0 to 1 (u^1/3 - u^4/) du then use the general power rule and plug in the numbers

[3/4u^4/3 - 3/7u^7/3] from 0to1

= [3/4(1)^4/3 -3/7(1)^7/3 - (0)]

=[3/4 - 3/7]

= 9/28

confidence assessment: 3

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19:55:48

What is the area?

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RESPONSE -->

9/28

confidence assessment: 3

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19:55:58

What integral did you evaluate to obtain the area?

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RESPONSE -->

previously answered

confidence assessment: 3

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19:56:06

What substitution did you use to evaluate the integral?

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RESPONSE -->

previously answered

confidence assessment: 3

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20:12:11

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

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RESPONSE -->

this problem isnt listed as being in the book, so i can't really get a good idea of what exactly it looks like.

confidence assessment: 0

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20:12:16

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

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RESPONSE -->

confidence assessment:

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20:12:23

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

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RESPONSE -->

confidence assessment:

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20:12:26

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

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RESPONSE -->

confidence assessment:

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20:13:20

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

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RESPONSE -->

The assignment was pretty difficult, i was able to figure them out by looking in the book. The last one i couldn't really understand the way it was set up. I wish i could have looked at it in the book.

confidence assessment: 3

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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&#This looks good. See my notes. Let me know if you have any questions. &#