Let f:R R be a twice differentiable function such that f(2)=1.

Question

Let $f:R	o R$ be a twice differentiable function such that $f(2)=1$. If $F(x)=xf(x)$ for all $x\in R$, $\int _{0}^{2}x{F}^{'}(x)dx=6$ and $\int _{0}^{2}{x}^{2}{F}^{''}(x)dx=40$, then ${F}^{'}(2)+\int _{0}^{2}F(x)dx$ is equal to :

Accepted Answer

$$\begin{aligned}
&\begin{aligned}
& \int_0^2 \mathrm{xF}^{\prime}(\mathrm{x}) \mathrm{dx}=6 \
& =\left.\mathrm{xF}(\mathrm{x})\right|_0 ^2-\int_0^2 \mathrm{f}(\mathrm{x}) \mathrm{dx}=6 \
& =2 \mathrm{~F}(2)-\int_0^2 \mathrm{xF}(\mathrm{x}) \mathrm{dx}=6[\therefore \mathrm{f}(2)=2 \mathrm{~F}(2)=2] \
& \int_0^2 \mathrm{xF}(\mathrm{x}) \mathrm{dx}=-2 \quad \ldots(1) \
& \Rightarrow \int_0^2 \mathrm{~F}(\mathrm{x}) \mathrm{dx}=-2\quad \ldots(2)
\end{aligned}\
\&\text { Also }\
\&\begin{aligned}
& \int_0^2 x^2 F^{\prime \prime}(x) d x=\left.x^2 F^{\prime}(x)\right|_0 ^2-2 \int_0^2 x F^{\prime}(x) d x=40 \
& =4 F^{\prime}(2)-2 \times 6=40 \
& F^{\prime}(2)=13 \
& \therefore F^{\prime}(2)+\int_0^2 F(x)=13-2=11
\end{aligned}
\end{aligned}$$

Solutions · 1