Assn 5

course mth 174

02-08-2008......!!!!!!!!...................................

01:17:54

what it is your antiderivative?

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

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01:18:03

Which formula from the table did you use?

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number 14

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01:21:59

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 and I derived p(x) untill i got a constant which was four times.

P(x)= x^4

P'(x)= 4x^3

P''(x)= 12x^2

P'''(x)= 24x

P''''(x)=24

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01:22:12

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

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-> questions

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01:23:03

What is your integral?

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ln(1z + (1/3z^3) + 8z) +c

The derivative of ln(z + (1 / 3 z^2) + 8 z) is (2z + 27)/(z(z + 27)).

The derivative of your result does not match the function.

You need to always take the derivative of your integral to see if you get the original integrand.

One solution to this problem:

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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01:38:39

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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26 the formula once expanded fit into the equation.

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01:41:36

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

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My partial fraction: (2y / y^3) - (2y / y^2) - (2y / y) - (1 / y)

The distributive law doesn't work this way. If it did, then

1 / (2 + 3) would be 1/2 + 1/3, which is equal to 5/6.

However 1 / (2 + 3) is 1/5.

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01:49:04

What is your result?

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ln l y-1 l + arctan - 1/2 ln l y^2 +1 l + C

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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01:53:18

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

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I used the formula given in the begining of the section to break the function down into different parts that could be more easily integrated.

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01:55:06

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

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

which would be equal to

(2y - 3 y^2 - 2 y^3 - 2 y^4) / y^3.

This is not equal to the original fraction.

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01:58:30

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

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-> questions

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02:15:26

What did you get for your integral?

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w = 1 - (z-1)^2

1 / 2z arctan z^2 / 2z +c

The derivative of this function does not give you the original integrand.

1 / 2z arctan z^2 / 2z means 1 / 2 * arctan(z)^2 / 2 * z; using grouping to emphasize the order of operations this would be (1 / 2) * ( ( arctan(z)^2) / 2 ) * z. I don't think this is what you intended.

More likely you mean

The derivative of 1 / (2z) * arctan (z^2) / (2z) is not equal to (z - 1) / sqrt( 2 z - z^2).

The derivative of 1 / (2z) * arctan (z^2) / (2z) is (z^2 - (z^4 + 1) arctan(z^2)) / (2·z^3 (z^4 + 1)).

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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02:16:05

What substitution did you use?

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w= 1 - (z- 1)^2

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02:18:44

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

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The once I can get the equations to fit into the formula for the integrals they are easily figuered out, but up untill that point the process is somewhat tedious..

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"

You had errors in most of these problems.

To check an integration you need to take the derivative of your result and see if it gives you the original integrand.

See my notes on this and on preceding Chapter 7 assignments. You should probably rework much of the work you have completed in Chapter 7. There is no doubt you can get this; the problem so far is that you appear to be oversimplifying the process of integration.

I'll be glad to answer specific questions if you do have difficulty.