Assignment 17

course mth 272

If your solution to a stated problem does not match the given solution, you should self-critique per instructions at

http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm.

Your solution, attempt at solution: If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.

017.

*********************************************

Question: `qQuery problem 6.2.21 (7th edition 6.3.18) integrate 3/(x ^ 2 - 3x)

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

3/ [ x(x-3) ] = A/x + B/(x-3)

3 = A(x-3) + B(x) or

3 = (A+B) x - 3 A

(A + B)x = 0

So A + B = 0

Solve for A and you get -1

Substitute this in to 3 = (A+B) x – 3A and b = 1

3 / (x^2 - 3x) = -1 / (x-3) + 1 / x.

ln | (x-3) / x | + c.

confidence rating #$&*3

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

`a First we factor x the demoninator:

3/ [ x(x-3) ]

Use partial fractions:

3/ [ x(x-3) ] = A/x + B/(x-3)

Multiply both sides by common denominator x(x-3) to get

3 = A(x-3) + B(x) or

3 = (A+B) x - 3 A, which is the same as

0 x + 3 = (A + B) x - 3 A.

The coefficients of x on both sides must be the same so we have

A + B = 0 (coefficients of x) and

-3 A = 3 .

From the second we get A = -1. Substituting this into the first we solve to get B = 1.

So our integrand is

3 / (x^2 - 3x) = -1 / (x-3) + 1 / x.

The integration is straightforward. We get

ln |x-3| - ln |x| + c , which we rewrite using the laws of logarithms as

ln | (x-3) / x | + c.

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary):

------------------------------------------------

Self-critique rating #$&*

*********************************************

Question: `qQuery problem 6.2.27 (7th edition 6.3.29) (was 6.3.27) integrate (x+2) / (x^2 - 4x)

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

(x+2)/x(x-4) = A/x + B/(x-4)

x+2= A(x-4) + Bx

x + 2 = (A+B)x – 4A

(A + B)x = x

So A + B = 1

-4 A = 2

A = -1/2

-1/2 + B = 1

B = 3/2

(-1/2) / x + (3/2) / (x-4)

3/2 ln |x-4| - 1/2 ln |x| + c

confidence rating #$&*3

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

`a We factor x out of the denominator to get

(x+2)/ [ x(x-4) ]

Use partial fractions:

(x+2)/x(x-4) = A/x + B/(x-4)

Multiply both sides by common denominator:

x+2 = A(x-4) + B(x) or

x+2 = (A+B) x - 4 A.

Thus

A + B = 1 and

-4 A = 2 so

A = -1/2 and

B = 3/2.

Our integrand becomes

(-1/2) / x + (3/2) / (x-4). The general antiderivative is easily found to be

3/2 ln |x-4| - 1/2 ln |x| + c, which can be expressed as

1/2 ln ( |x-4|^3 / | x | ) + c

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary):

------------------------------------------------

Self-critique rating #$&*

*********************************************

Question: `qQuery Add comments on any surprises or insights you experienced as a result of this assignment.

After doing the problems and watching the videos I now understand how to do these types of problems

"

&#This looks good. Let me know if you have any questions. &#