Calculus problem

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e^1

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Calculus problem

« on: November 05, 2012, 10:56:19 pm »

0

Question:

If $\frac{4}{3} \int (\frac{x - a}{x^2 - 3x}) = \frac{2}{3} log_{e}(x^2 - 3x)$ , find the value of $a$ .

Edit: Thanks b^3 for the clear answer!

« Last Edit: November 05, 2012, 11:06:48 pm by xp3r009 »

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b^3

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Re: Calculus problem

« Reply #1 on: November 05, 2012, 11:02:46 pm »

+6

$\begin{alignedat}{1}\frac{4}{3}\int\frac{x-a}{x^{2}-3x}dx & =\frac{2}{3}\log_{e}\left(x^{2}-3x\right)<br />\\ \frac{4}{3}\frac{x-a}{x^{2}-3x} & =\frac{d}{dx}\left(\frac{2}{3}\log_{e}\left(x^{2}-3x\right)\right)<br />\\ \frac{4}{3}\frac{x-a}{x^{2}-3x} & =\frac{2}{3}\frac{1}{x^{2}-3x}\times2x-3<br />\\ \frac{4}{3}\frac{x-a}{x^{2}-3x} & =\frac{4}{3}\frac{x-\frac{3}{2}}{x^{2}-3x}<br />\\ \therefore a & =\frac{3}{2}<br />\end{alignedat}$

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Moko

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Re: Calculus problem

« Reply #2 on: November 05, 2012, 11:03:56 pm »

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a=3/2....this is a spesh question, I'd explain it but I still haven't learnt LaTex..Until I do, it would probably be the end of the methods exams lol.

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Moko

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Re: Calculus problem

« Reply #3 on: November 05, 2012, 11:05:09 pm »

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Oh ok b^3's got it...if you do it that way its methods...but if you do it via integration then it would be spesh.

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