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QCE Stuff => QCE Mathematics Subjects => QCE Subjects + Help => QCE Essential Mathematics => Topic started by: scheely on February 24, 2021, 04:43:41 pm

Title: anti-derivative
Post by: scheely on February 24, 2021, 04:43:41 pm: Hi - was looking for help in regards determining anti-derivatives.
I am asked to find a derivative of f(x) = x^2ln(2x), finding f'(x); and then hence the anti derivative integer xln(2x)dx. (where ln is log e).
Title: Re: anti-derivative
Post by: fun_jirachi on February 24, 2021, 05:03:30 pm: Welcome to the forums! :D

Now, the derivative of $x^2 \ln (2x) = x^2 \times \frac{2}{2x} + 2x\ln (2x) = x + 2x\ln (2x)$ by the product rule.

Hence, the antiderivative of $x\ln (2x)$ will be equal to $\frac{1}{2}\int x + 2x\ln (2x) - x \ dx$.
Evaluating this integral now:
$\begin{align*}\frac{1}{2}\int x + 2x\ln (2x) - x \ dx &= \frac{1}{2}\int x + 2x\ln (2x) \ dx - \frac{1}{2}\int x \ dx<br />\\&= \frac{1}{2} x^2 \ln (2x) - \frac{x^2}{4} + C <br />\end{align*}$

Note how we use the fundamental theorems of calculus and our previous work + how we manipulate the expression provided in the question into a more familiar form that allows us to employ said method. Try a few more of these and it should bed in pretty handily.

Hope this helps :)