Assignment 5 Query

course Mth 174

Sorry about getting this in late. I am planning on taking the test Monday morning first thing. Quick question: Will I need the tables of indefinite integrals for Test 1?

Nothing on Test 1 requires tables.

Test 2 has problems that do require a table, and a table is included on that test.

I believe Test 1 goes through Section 7.2.

Thanks!

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assignment #005

VJ]

Physics II

09-27-2007

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21:36:22

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

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RESPONSE -->

Using the tables, this is what I found:

Indefinite integral = (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) + C

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21:36:27

what it is your antiderivative?

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See prior answer.

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21:36:51

Which formula from the table did you use?

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I used formula III.14.

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21:37:54

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

I used four derivatives of p(x):

p'(x) = 4x^3

p''(x) = 12x^2

p'''(x) = 24x

p''''(x) = 24

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21:39:18

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

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I found the derivative to be:

arctan(z+2) +C

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21:39:30

What is your integral?

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Well, the integral is listed as above.

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21:43:04

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 didn't. I used a substitution:

u = z + 2

Which left the equation:

1 / (1 + u^2)

I recognized this to be the derivative of the arctan function, and so I found:

arctan(u) + C

Putting the original value of u in:

arctan(z + 2) + C

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22:29:46

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

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RESPONSE -->

I think I found an answer to this one. The functions I found are as follows:

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

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22:29:52

What is your result?

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See above.

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22:37:35

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

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RESPONSE -->

Looking for common factors doesn't work here, so I had to use for quadratic-style factoring. I looked at the factors of the highest power function:

y * y^2

And the factors of the last number, -1:

-1 * 1

So I can set those into two possible formats:

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

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

The simplest way I know to find out which one fits is to just multiply it through.

Any suggestions on this account would be greatly appreciated, as I am sure this isn't the best way.

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22:38:14

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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I had to use the form:

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

You've got most of it.

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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22:58:15

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

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RESPONSE -->

I'm guessing I really should wait for the next section to input my answer.

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23:30:44

What did you get for your integral?

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Initially I only worked on finding a possible substitution by going to the inmost function and completing the square. My result:

w = 1 - (z - 1)^2

But I am not completely sure where to go from there. Using substition, here are my steps:

dw = 2(z -1) dz

dz = (1/2) dw / (z - 1)

Substituting into the original equation:

= (z - 1) / 'sqrt (w) * (1/2) dw / (z - 1)

= (1/2) dw / 'sqrt (w)

Finding the antiderivative of this:

antiderivative = 'sqrt (w)

Expanded:

= 'sqrt ( 1 - (z - 1)^2 ) + C

Is that right?

That works. But there's a simpler way to get there:

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).

This is equivalent to your answer, since 1 - (z - 1)^2 = 1 - (z^2 - 2z + 1) = 1 - z^2 + 2 z - 1 = 2 z - z^2.

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23:30:51

What substitution did you use?

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See prior answer.

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23:32:19

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

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To be answered in the next box...

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23:41:43

What is your integral and how did you obtain it?

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I found the integral to be:

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

I obtained it by first factoring the denominator, 2y^2 +3y +1:

(y + 2) / ( (2y + 1)(y+1) )

and then splitting it up into partial fractions:

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

Solving for A and B:

A = 3

B = -1

This gave me the result:

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

I then found the antiderivative to be:

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

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23:42:50

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

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RESPONSE -->

Whew. That was wild. But it was still incredibly interesting to see how substition can be used to chain derivatives and antiderivatives together until a complex equation is simple, and then simply reverse the process to find the answer.

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Excellent work. See my notes.

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Let me know if you have questions. &#