Question

Let $f:\mathrm{ℝ}\to \mathrm{ℝ}$ be a differentiable function such that its derivative $f\text{'}$ is continuous and $f\left(\mathrm{\pi }\right)=-6$

If $F:\left[0,\mathrm{\pi }\right]\to \mathrm{ℝ}$ is defined by $F\left(x\right)={\int }_0^{x}f\left(x\right)dx$, and if ${\int }_0^{\mathrm{\pi }}\left(f\text{'}\left(x\right)+F\left(x\right)\right)\cos xdx=2$, Then the value of $f\left(0\right)$_____

Answer 1

Finding $f\left(0\right)$:

$F\left(x\right)={\int }_0^{x}f\left(x\right)dx$

By newton Leibniz rule:

$F\text{'}\left(x\right)=f\left(x\right)$

We are given that

$$\begin{gathered} {\int }_0^{\mathrm{\pi }}\left(f\text{'}\left(x\right)+F\left(x\right)\right)\cos xdx=2 \\ \Rightarrow {\int }_0^{\mathrm{\pi }}\left(f\text{'}\left(x\right)+F\left(x\right)\right)\cos xdx={\int }_0^{\mathrm{\pi }}f\text{'}\left(x\right)\cos x+{\int }_0^{\mathrm{\pi }}F\left(x\right)\cos xdx \end{gathered}$$

Using integration by parts we get on

$$\begin{gathered} {\int }_0^{\mathrm{\pi }}f\text{'}\left(x\right)\cos xdx+{\left[F\left(x\right)\sin x\right]}_0^{\mathrm{\pi }}-{\int }_0^{\pi }F\text{'}\left(x\right)\sin xdx=2 \\ \Rightarrow {\int }_0^{\mathrm{\pi }}f\text{'}\left(x\right)\cos xdx-{\int }_0^{\mathrm{\pi }}f\left(x\right)\sin xdx=2 \\ \Rightarrow {\int }_0^{\mathrm{\pi }}\frac{d}{dx}\left(f\left(x\right)\cos x\right)=2 \\ \Rightarrow {\left[f\left(x\right)\cos x\right]}_0^{\mathrm{\pi }}=2 \\ \Rightarrow f\left(\mathrm{\pi }\right)\left(-1\right)-\left(\mathrm{f}\left(0\right)\right)=2 \\ \Rightarrow 6-\mathrm{f}\left(0\right)=2 \\ \Rightarrow \mathrm{f}\left(0\right)=4 \end{gathered}$$

Hence the value of f(0) is 4