Question

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

Answer 1

Answer: 4.00

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