course Mth 272
??????Having hard time following the querry and the sections. They seem to be off from the eight edition.
017.
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Question: `qQuery problem 6.3.18 integrate 3/(x ^ 2 - 3x)
Your solution
3 / (x^2 3x) - factor out x = 3 / x(x-3)
3 / (x^2 3x) = (A / x-3) + (B / x))
3 = A(x-3) + Bx
To Solve for B we let x = 3
3 = A(3-3) + B(3) = 3 = A(0) + 3B
B = 1
To Solve for A we let x = 0
3 = A(0-3) + B(0)
3 = -3A
-1 = A
So we have 3 / x^2 3x dx = integral [(-1 / x -3) + 1 / x] dx
= ln |x-3| - ln|x| + c
confidence rating #$&*2
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Given Solution:
`a First we factor x the demoninator:
3/ [ x(x-3) ]
Use partial fractions:
3/ [ x(x-3) ] = A/x + B/(x-3)
Multiply both sides by common denominator x(x-3) to get
3 = A(x-3) + B(x) or
3 = (A+B) x - 3 A, which is the same as
0 x + 3 = (A + B) x - 3 A.
The coefficients of x on both sides must be the same so we have
A + B = 0 (coefficients of x) and
-3 A = 3 .
From the second we get A = -1. Substituting this into the first we solve to get B = 1.
So our integrand is
3 / (x^2 - 3x) = -1 / (x-3) + 1 / x.
The integration is straightforward. We get
ln |x-3| - ln |x| + c , which we rewrite using the laws of logarithms as
ln | (x-3) / x | + c.
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Self-critique (if necessary):
Understand the problem just forgot that I could rewrite the integral.
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Self-critique rating #$&*3
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Question: `qQuery problem 6.3.29 (was 6.3.27) integrate (x+2) / (x^2 - 4x)
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
Fist we factor out an x = x + 2 / (x(x-4))
x + 2 / (x^2 4x) = A(x-4) + B(x)
Right idea, but you would need to write this as
(x + 2) / (x^2 4x) = A/(x-4) + B/(x)
in order for the denominators on the two sides to match.
To solve for B we let x = 4
4 + 2 = A(4 4) + 4B
6 = 4B
B = 3/2
Letting x = 4 would be a good thing to try in solving your equation for B. However if you correct your equation, this won't work, since you would end up with 0 in the denominator of the term A ( x - 4 ).
In your equation, in any case, substituting x = 4 would lead to 0 in the denominator of the left-hand side: (x + 4) / (x^2 - 4 x) would be 4 * 4 / (4^2 - 4 * 40 = 16 / 0, which is undefined. So the good idea of letting x = 4 doesn't pan out in this case.
To Solve for A we let x = 0
0 + 2 = A(0-4) + B(0)
2 = -4A
A = -1/2
We have than [x + 2 / (x^2 4x)] dx = [ -1/2 / x-4 + 3/2 / x ] dx
= -1/2 ln |x-4| + 3/2 ln |x| + c
confidence rating #$&*2
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution:
`a We factor x out of the denominator to get
(x+2)/ [ x(x-4) ]
Use partial fractions:
(x+2)/x(x-4) = A/x + B/(x-4)
Multiply both sides by common denominator:
x+2 = A(x-4) + B(x) or
x+2 = (A+B) x - 4 A.
Thus
A + B = 1 and
-4 A = 2 so
A = -1/2 and
B = 3/2.
Our integrand becomes
(-1/2) / x + (3/2) / (x-4). The general antiderivative is easily found to be
3/2 ln |x-4| - 1/2 ln |x| + c, which can be expressed as
1/2 ln ( |x-4|^3 / | x | ) + c
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Self-critique (if necessary):
Though I did it write not sure why I did not get the answer.
There are a couple of errors in your solution, though your overall approach is sound. See my notes.
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Self-critique rating #$&*
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Question: `qQuery Add comments on any surprises or insights you experienced as a result of this assignment
Having a hard time following the querry and sections on chapter 6. They dont go along with the eight edition of the book
I believe I've straightened out the assignments, at least as far as getting the right section with the right query for the 8th edition. You might need to refresh the Assignments page for the 8th edition. I'll be taking another look at the correspondences tomorrow, as well as the problem numbers, to eliminate as many errors as possible.
&#Be sure to see my note(s), inserted at various places in this document, and let me know if you have questions. &#