Assignment 5

course MTH 174

?Pio??????}?assignment #005

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Physics II

11-20-2007

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22:40:59

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

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RESPONSE -->

integral of x^4 e^(3x) =

(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) =

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

Applied formula III-14 from the back of the book.

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22:41:14

what it is your antiderivative?

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RESPONSE -->

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

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22:41:32

Which formula from the table did you use?

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RESPONSE -->

Formula III-14

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22:44:05

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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RESPONSE -->

a = 3

p(x) = x^4

p'(x) = 4x^3

p''(x) = 12x^2

p'''(x) = 24x

p''''(x) = 24

Four derivatives gave a total of 5 terms, including the original p(x).

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23:53:21

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

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RESPONSE -->

integral of 1 / [ 1 + (z+2)^2 ) ] taken with respect to z.

w = (z + 2)

integral of 1 / (1^2 + w^2) dw = (1 / `sqrt(1)) * arctan (w / `sqrt(1)) + C =

(1 / `sqrt(1)) * arctan ((z + 2 ) / `sqrt(1)) + C =

arctan (z+ 2) + C

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23:53:53

What is your integral?

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RESPONSE -->

integral of 1 / [ 1 + (z+2)^2 ) ] taken with respect to z.

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23:56:02

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 -->

I used substitution so that the equation would fit formula V-24.

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18:40:40

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

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RESPONSE -->

A partial fractions decomposition for 2y / ( y^3 - y^2 + y - 1) is

2y = (Ay + b) / (y^2 + 1) + C / (y - 1)

A = -1, B = 1, C = 1

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

leaves us the integral of

(-y) / (y^2 + 1) + 1 / (y^2 + 1) + 1 / (y - 1) =

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

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18:45:18

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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RESPONSE -->

(Ax + B) / (y^2 + 1) + C / (y-1)

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11:56:47

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

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RESPONSE -->

Indefinite Integral of (z-1)/`sqrt(2z-z^2) * dz

let u = 2z - z^2

du = 2 - 2z * dz

-1 / 2 du = z - 1 * dz

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

-1 / 2 * indefinite integral of 1 / `sqrt(u) * du =

-1 / 2 * indefinite integral of u^(-1/2) du =

(-1 / 2) * 2(u)^(1 / 2) = - `sqrt (u) + C =

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

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11:57:05

What did you get for your integral?

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RESPONSE -->

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

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11:57:31

What substitution did you use?

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RESPONSE -->

let u = 2z - z^2

du = 2 - 2z * dz

-1 / 2 du = z - 1 * dz

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13:05:44

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

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RESPONSE -->

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

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

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

ind. int. of -1 / (2y + 1) + ind. int. of 1 / (y + 1) + ind. int. of 4 / (2y + 1) + ind. int. of. -2 / (y + 1) =

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

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13:07:15

What is your integral and how did you obtain it?

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RESPONSE -->

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

obtained by partial fractions decomposition,

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

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

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

ind. int. of -1 / (2y + 1) + ind. int. of 1 / (y + 1) + ind. int. of 4 / (2y + 1) + ind. int. of. -2 / (y + 1) =

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

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13:08:15

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

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RESPONSE -->

I've developed a method of approach for solving those integrals.

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Excellent work.