Assignment 5

course Mth 174

???}????v??~q?assignment #005???g??x??m??????Physics 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.

02-03-2007

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19:44:45

Query problem 7.3.17 (3d edition #15) x^4 e^(3x)

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19:51:54

what it is your antiderivative?

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Use #14 p(x) = x^4 a=3

Antiderivative of x^4e^(3x) dx =

(1/3)x^4e^(3x)-(1/9)(4x^3)e^(3x)+(1/27)(12x^2)e^(3x)-(1/81)(24x)e^(3x)+(1/243)(24)e^(3x)+C

=(1/3)x^4e^(3x) - (4/9)x^3e^(3x) + (4/9)x^2e^(3x) - (8/27)xe^(3x) + (8/81)e^(3x) + C

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19:52:03

Which formula from the table did you use?

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#14

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19:54:02

You should have used formula 14. What was your value of a, what was p(x), how many derivatives did you use and what were your derivatives of p(x)?

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See previous response.

4 derivatives: p'(x) = 4x^3 p''(x) = 12x^2 p'''(x) = 24x p''''(x) = 24

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19:54:33

Query problem 7.3.33 (3d edition #30) 1 / [ 1 + (z+2)^2 ) ]

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19:59:18

What is your integral?

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u = z+2 du = dz

Integral of 1/[1+(z+2)^2]

= Integral of 1/(1+u^2) du

Use #24

a=1

= (1/1) arctan (u/1) + C

= arctan u + C

= arctan (z+2) + C

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19:59:45

Which formula from the table did you use and how did you get the integrand into the form of this formula?

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RESPONSE -->

See previous response

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20:16:35

7.4.6 (#8 in 3d edition). Integrate 2y / ( y^3 - y^2 + y - 1)

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20:27:26

What is your result?

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Integral of 2y/(y^3-y^2+y-1) dy

Factor denominator

= Integral of 2y/[(y^2+1)(y-1)] dy

Use partial fractions

= Integral of (Ay+B)/(y^2+1) + C/(y-1) dy

Creating common denominator and multiplying to find numerators:

2y = Ay^2+By-Ay-B+Cy^2+C

Thus:

A+C = 0 B-A = 2 -B+C = 0

Solving: A = -1 B = 1 C = 1

Substituting in integral:

= Integral of (-y+1)/(y^2+1) dy + Integral of 1/(y-1) dy

Use #25 on first integral, a = 1, b = -1, c = 1

= (-1/2)ln |y^2+1| + (1/1) arctan (y/1)

= (-1/2)ln |y^2+1| + arctan y

Back to full integral

= (-1/2)ln |y^2+1| + arctan y + ln |y-1| + C

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20:27:36

How did you factor your denominator to get the integrand into a form amenable to partial fractions?

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RESPONSE -->

See previous response

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20:27:56

After the integrand was in the form 2y / [ (y-1) * (y^2 + 1) ], what form of partial fraction did you use?

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See previous response

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20:50:36

7.4.29 (3d edition #24). Integrate (z-1)/`sqrt(2z-z^2)

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20:59:17

What did you get for your integral?

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Integral of (z-1)/[sqrt (2z-z^2)] dz

Complete the square:

= Integral of (z-1)/[sqrt(1-(z-1)^2)] dz

Substitute: u = (1-(z-1)^2) du = (2-2z) -du/2 = (z-1)dz

= Integral of 1/sqrt u * -du/2

= (-1/2) * Integral of 1/sqrt u du

= (-1/2)(2) u^(1/2) + C

= -[u^(1/2)] + C

= - sqrt [1-(z-1)^2] + C

= - sqrt (2z-z^2) + C

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20:59:27

What substitution did you use?

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See previous response

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21:27:39

7.4.40 (3d edition #28). integrate (y+2) / (2y^2 + 3y + 1)

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21:37:52

What is your integral and how did you obtain it?

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Integral of (y+2)/(2y^2+3y+1) dy

Factor denominator:

= Integral of (y+2)/[(y+1)(2y+1) dy

Use partial fractions:

- Integral of A/(y+1) + B/(2y+1) dy

Create common denominator, multiply across, set numerators equal:

y+2 = A(2y+1) + B(y+1)

y+2 = 2Ay+A+By+B

Solving: A = -1, B = 3

Subsitute in integral:

= Integral of -1/(y+1) dy + Integral of 3/(2y+1) dy

= - Integral of 1/(y+1) dy + 3*Integral of 1/(2y+1) dy

First integral = - ln |y+1|

Second integral:

Substitute u = 2y du = 2dy du/2 = dy

= 3* Integral of 1/(u+1) du/2

= (3/2) * Integral of 1/(u+1) du

= (3/2) ln |u+1|

Substitute back:

= (3/2) ln |2y+1|

Combining integrals:

= - ln |y+1| + (3/2) ln |2y+1| + C

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21:42:30

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

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There are sometimes deadends, particularly when using trig substitutions, that require going back and trying something else. Number 7.4-29, in particular, didn't work according to the book formula (example 12), which suggests using u = z-1. This didn't work out, but trying u = 1-(z-1)^2 did.

On #29 if you let u = 2z - z^2, then you get du = 2 - 2 z, which is 2 ( 1- z).

So the numerator 1 - z is double u ' and your integrand becomes 1 / (2 sqrt(u) ). The antiderivative is just sqrt(u), which becomes sqrt(2 z - z^2).

This is much simpler than your substitution, which however was still very good--it's not easy to find the expression you used.

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Very good. Be sure to see my note in response to your final note above. Let me know if you have questions.