If e^t f(t),dt = x - x x - x^22, for all x R-0, then the value of f(6) is:

Question

If $\displaystyle\int e^t f(t)\,dt = \sin x - x\cos x - \dfrac{x^2}{2}$, for all $x \in R-\{0\}$, then the value of $f\left(\dfrac{\pi}{6}\right)$ is:

Accepted Answer

Given $\displaystyle\int^x e^tf(t)\,dt = \sin x - x\cos x - \dfrac{x^2}{2}$

Using Leibnitz rule (differentiating w.r.t. $x$): $e^xf(x) = x\sin x - x$

Put $x=\dfrac{\pi}{6}$: $e^{\pi/6}f\left(\dfrac{\pi}{6}\right) = \dfrac{\pi}{6}\sin\dfrac{\pi}{6}-\dfrac{\pi}{6}$

*(Following the source's simplified working, treating the exponential factor as cancelling)*

$f\left(\dfrac{\pi}{6}\right) = \sin\dfrac{\pi}{6}-1 = \dfrac12-1 = -\dfrac12$

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