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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematical Methods CAS => Topic started by: e^1 on November 05, 2012, 10:56:19 pm

Title: Calculus problem
Post by: e^1 on November 05, 2012, 10:56:19 pm: 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!
Title: Re: Calculus problem
Post by: b^3 on November 05, 2012, 11:02:46 pm: $\begin{alignedat}{1}\frac{4}{3}\int\frac{x-a}{x^{2}-3x}dx & =\frac{2}{3}\log_{e}\left(x^{2}-3x\right) \\ \frac{4}{3}\frac{x-a}{x^{2}-3x} & =\frac{d}{dx}\left(\frac{2}{3}\log_{e}\left(x^{2}-3x\right)\right) \\ \frac{4}{3}\frac{x-a}{x^{2}-3x} & =\frac{2}{3}\frac{1}{x^{2}-3x}\times2x-3 \\ \frac{4}{3}\frac{x-a}{x^{2}-3x} & =\frac{4}{3}\frac{x-\frac{3}{2}}{x^{2}-3x} \\ \therefore a & =\frac{3}{2} \end{alignedat}$
Title: Re: Calculus problem
Post by: Moko on November 05, 2012, 11:03:56 pm: 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.
Title: Re: Calculus problem
Post by: Moko on November 05, 2012, 11:05:09 pm: 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.