4
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I used partial fractions to get:

$\displaystyle\frac{A}{x-4}+\frac{B}{x-2}= \frac{x-6}{(x-4)(x-2)}$

$A = -1$

$B = 2$

$\displaystyle\int_{0}^1\frac{-1}{x-4}+\frac{2}{x-2}dx$

Found the anti-derivative to be:

$(-\ln|x-4| + 2 \ln|x-2|)_0^{1}$

My answer came out to be around -1.1, what am I doing wrong? Thanks for any help.

Edit- Apparently WebAssign thinks that I'm wrong for some reason: enter image description here

  • 0
    @mathisnotmyforte I find it really disappointing that you choose/prefer $-1.1$ as an answer rather than $-\log 3$.2012-03-06

3 Answers 3

0

In the interval $[0,1]$ we have $\frac{-5}{3}\leq\frac{x-6}{x^2-6x+8}\leq\frac{-3}{4}$, hence:

$ \frac{-5}{3}\leq\int_0^1\frac{x-6}{x^2-6x+8}\leq\frac{-3}{4} $

3

You're doing nothing wrong.

Simply substitute the boundary numbers, to get

$A = -\log|1-4|+2\log|1-2| + \log|-4|-2\log|-2|$

$A = -\log3+2\log1 + \log4-2\log2$

$A = -\log3 \approx-1.09861$

1

-1.1 is correct according to Wolfram.

http://www.wolframalpha.com/input/?i=integrate+%28x-6%29%2F%28x%5E2-6x%2B8%29+dx+from+0+to+1