Question

Let $f: R \rightarrow R$ be a differentiable function such that its derivative $f^{\prime}$ is continuous and $f(\pi)=-6$.
If $F$ : $[0$, $\pi] \rightarrow R$ is defined by $F(x)=\int\limits_{0}^{x} f(t) d t$, and if $\int\limits_{0}^{\pi}\left(f^{\prime}(x)+F(x)\right) \cos x d x=2$
then the value of $f (0)$ is _____

Answer 1

$\int\limits_{0}^{\pi}\left(f^{\prime}(x)+F(x)\right) \cos x d x$
$\int\limits_{0}^{\pi}\left[F^{\prime \prime}(x) \cos x-F^{\prime}(x) \sin x+F^{\prime}(x) \sin x+F(x) \cos x\right] d x$
$=\int\limits_{0}^{\pi}\left(F^{\prime}(x) \cos x+F(x) \sin x\right)$
$=-F^{\prime}(\pi)-F^{\prime}(0)=-f^{\prime}(\pi)-f(0)=2$
$ f(0)=4$