Assignment 5

course Mth 174

Sorry for not turning in any work for the past week, I was on vacation. I will take my first test later this week and then use the following week to get caught back up.Also, I was wondering if you have been getting my emails about last semester's grade. My parents pay for my classes and they have started pecking at me about it.

I thought I had replied about the grade. I'll be out of town until Friday; email me then and I can give you the details.

assignment #005

V˪˵}Ӿ

Physics II

06-23-2008

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15:56:34

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

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I set x^4 = p(x)

3 = a

I also used the table, III number 14.

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

=

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

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15:56:37

what it is your antiderivative?

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15:56:46

Which formula from the table did you use?

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Part III, number 14.

int(p(x) e^ax dx = (1/a) p(x) e^ax - (1/a) int(p'(x) e^ax dx

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15:56:57

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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a = 3

p(x) = x^4

4 derivatives

x^4

4x^3

12x^2

24x

24

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15:57:13

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

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I used part V number 24 to solve this.

int(1 / (x^2 + a^2)) = (1/a) arctan (x/a)

x^2 = 1 or 1^2

a^2 = (z + 2)^2

(1/(z+2)) arctan (1/(z + 2)) + C

z is the variable of integration in the given problem, x is the variable of integration in the table. a is a constant, so a won't be z + 2.

The antiderivative of 1 / (a^2 + x^2) is 1/a * arcTan(x/a).

Unlike some formulas in the table, this formula is easy to figure out:

1 / (a^2 + x^2) = 1 / (a^2( 1 + x^2/a^2) ) = 1/a^2 ( 1 / (1 + (x/a)^2).

Let u = x / a so du = dx / a; you get integrand 1 / a * (1 / (1 + u^2) ) , which has antiderivative 1/a * arcTan(u) = 1/a * arcTan(x/a).

You don't really need to know all that, but it should clarify what is constant and what is variable.

Starting with int(1/ (1+(z+2)^2) dz ), let x = z + 2 so dx = dz. You get

int(1/ (1+x^2) dx ), which is formula 24 with a = 1. The result is

1/1 * arcTan(x/1), or just arcTan(x). Since x = z + 2, the final form of the integral is

arcTan(z+2).

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15:57:14

What is your integral?

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15:57:24

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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Part V, number 24

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15:57:40

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

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I used Partial Fractions.

I broke it down to 2y / [(y-1)(y^2 + 1)]

2y / [(y-1)(y^2 + 1)] = (Ay + B) / (y^2 + 1) + C / (y - 1)

I multiplied both sides by (y-1)(y^2 + 1) to get 2y = (Ay + B)(y - 1) + C(y^2 + 1)

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

= (A + C)y^2 + (-A + B)y + (-B + C)

Then I got a little confused.

(A + C)y^2 = 2y

(-A + B)y = 2y

(-B + C) = 2y

I won't go through the process of solving for anything on this answer, although I did it on paper. I became stuck after this.

The denominator factors by grouping:

y^3 - y^2 + y 1 = (y^3 + y) (y^2 + 1) = y ( y^2 + 1) 1 ( y^2 + 1) = (y 1) ( y^2 + 1).

Using partial fractions you would then have

(a y + b) /(y^2 + 1) + c /(y-1) = y / ( (y^2+1)(y-1)) where we need to evaluate a, b and c.

Putting the left-hand side over a common denominator (multiply the first fraction by (y-1)/(y-1) and the second by (y^2+1) / (y^2 + 1)) we obtain:

[ (a y + b)(y-1) + c(y^2+1) ] / ((y^2+1)(y-1)) = y / ((y^2+1)(y-1)). The denominators are identical so the numerators are equal, giving us

(a y + b)(y-1) + c(y^2+1) = y, or

a y^2 + (-a + b) y - b + c y^2 + c = y. Grouping the left-hand side:

(a + c) y^2 + (a - b) y + c - b = y. Since this must be so for all y, we have

a + c = 0 (this since the coefficient of y^2 on the right is 0)

-a + b = 1 (since the coefficient of y on the right is 1)

c - b = 0 (since there is no constant term on the right).

From the third equation we have b = c; from the first a = -c. So the second equation

becomes

c + c = 1, giving us 2 c = 1 and c = 1/2.

Thus b = c = 1/2 and a = -c = -1/2.

Our integrand (a y + b) /(y^2 + 1) + c /(y-1) becomes

(-1/2 y + 1/2 ) / (y^2 + 1) + 1/2 * 1 / (y-1), or

-1/2 y / (y^2 + 1) + 1/2 * 1 / (y^2 + 1) + 1/2 * 1 / (y-1).

An antiderivative is easily enough found with or without tables to be

-1/2 ln(y^2 + 1) + 1/2 arcTan(y) + 1/2 ln | y - 1 | + c.

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15:57:54

What is your result?

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15:58:05

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

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My denominator was (y - 1)(y^2 + 1)

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15:58:51

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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2y / [(y-1)(y^2 + 1)] = (Ay + B) / (y^2 + 1) + C / (y - 1)

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15:59:05

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

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To complete the square on in the denominator I rewrote it as z^2 - 2z + 0

I then completed the square to get (z^2 - 2z + 1) - 1

I then factored it to get (z - 1)^2 - 1 = 0

or 0 = 1 - (z - 1)^2

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

I subsituted x - 1 = sin(z)

w = sin(z) w' = cos(z)

(z - 1) / sqrt[1 - (sin(z))^2]

The Pythagorean identity, cos^2 + sin^2 = 1

so 1 - sin^2 = cos^2

[(z-1) / sqrt(cos^2(z))] * cos(z)

int( (z - 1)(cos(z) / cos(z)

z - 1 + C

If you let u = 2z - z^2 you get du = (2 - 2z) dz or -2(z-1) dz; thus (z-1) dz = -du / 2.

So (z-1) / `sqrt(2z - z^2) dz = 1 / sqrt(u) * (-du/2) = - .5 u^-.5 du, which integrates to

-u^.5. Translated in terms of the original variable z we get

-sqrt(2z-z^2).

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15:59:10

What did you get for your integral?

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15:59:11

What substitution did you use?

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16:20:06

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

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int( (y+2) / (2y^2 + 3y + 1) dy

int( (y+2) / (2y + 1)(y + 1) dy

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

Multiply by (2y + 1)(y + 1)

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

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

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

(A + 2B)y = y

A = 2B

A + B = 2

2B + B = 2

3B = 2

B = 2/3

A = 2(2/3) = 4/3

(y+2) / (2y^2 + 3y + 1) = (4/3) / (2y + 1) + (2/3) / (y + 1)

This is as far as I know to go.

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16:20:08

What is your integral and how did you obtain it?

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16:20:42

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

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