Assignment 5

course Mth 174

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assignment #005

?????S??B+???Physics II

02-08-2009

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08:47:56

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

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RESPONSE -->

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

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08:48:01

what it is your antiderivative?

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RESPONSE -->

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

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08:53:24

Which formula from the table did you use?

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RESPONSE -->

III 14 which is integral p(x) e^(ax) dx

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08:57:28

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^3

I used 4 derivatves:

p'(x) = 3x^2

p''(x) = 6x

p'''(x) = 6

p''''(x) = 0

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08:58:24

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

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RESPONSE -->

arctan (z + 2) + C

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08:58:31

What is your integral?

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RESPONSE -->

arctan (z + 2) + C

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09:03:26

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 -->

V 24 which is the integral 1/(x^2 + a^2)dx = 1/a arctan x/a + C

I completed the squar of the denomenator in order to get teh integrand into the form of this formula

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09:11:35

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

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RESPONSE -->

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

A / (y^3 - y^2) + B / (y-1)

Right idea. The details:

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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09:11:39

What is your result?

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RESPONSE -->

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09:13:10

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

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RESPONSE -->

if you factor a y out of the equation you get 1/ (y^2 - y)

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09:14:06

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 -->

How did you get this form of the integrand?

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09:14:56

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

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RESPONSE -->

w = 1- (z-1) ^2

1 - (z-1)^2 is a second-degree polynomial, i.e., a quadratic function. Its derivative is a first-degree polynominal, a linear function. The given integrand is not a linear function, so your result doesn't check out.

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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09:15:00

What did you get for your integral?

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RESPONSE -->

w = 1- (z-1) ^2

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09:18:10

What substitution did you use?

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RESPONSE -->

I used tangent substitution

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09:41:56

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

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RESPONSE -->

I used V - 27

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

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

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09:42:01

What is your integral and how did you obtain it?

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RESPONSE -->

I used V - 27

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

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

(y+2) / (2y^2 + 3 y + 1) =
(y + 2) / ( (2y + 1) ( y + 1) ) =
(y + 2) / ( 2(y + 1/2) ( y + 1) ) =
1/2 * (y + 2) / ( (y + 1/2) ( y + 1) )
The expression
(y + 2) / ( (y + 1/2) ( y + 1) )
is of the form
(cx + d) / ( (x - a)(x - b) )
with c = 1, d = 2, a = -1/2 and b = -1.
Its antiderivative is given as
1 / (a - b) [ (ac + d) ln | x - a | - (bc + d) ln | x - b | ] + C.
The final result is obtained by substitution.

Assignment 5

You had some good solutions, and some errors.

&#Let me know if you have questions. &#