Determination of Function Question - None

Question

$\int f'(x)dx = x \cos \pi x$ and $f(0) = 0$ then value of $f\left(\frac{1}{3}\right)$

Accepted Answer

$f(x) + C = x \cos \pi x$
$f(0) + C = 0 \cos 0$
$0 + C = 0$
$ herefore C = 0$
$ herefore f(x) = x \cos \pi x$
$f\left(\frac{1}{3} ight) = \frac{1}{3} \cos \frac{\pi}{3} \Rightarrow \frac{1}{3} imes \frac{1}{2} = \frac{1}{6}$

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