Let $\mathrm{f}: \mathbf{R} \rightarrow \mathbf{R}$ be a twice differentiable function such that $f(2)=1$. If $\mathrm{F}(x)=x f(x)$ for all $x \in \mathbf{R}$, $\int_0^2 x \mathrm{~F}^{\prime}(x) \mathrm{d} x=6$ and $\int_0^2 x^2 \mathrm{~F}^{\prime \prime}(x) \mathrm{d} x=40$, then…

Question

Let $\mathrm{f}: \mathbf{R} \rightarrow \mathbf{R}$ be a twice differentiable function such that $f(2)=1$. If $\mathrm{F}(x)=x f(x)$ for all $x \in \mathbf{R}$, $\int_0^2 x \mathrm{~F}^{\prime}(x) \mathrm{d} x=6$ and $\int_0^2 x^2 \mathrm{~F}^{\prime \prime}(x) \mathrm{d} x=40$, then $\mathrm{F}^{\prime}(2)+\int_0^2 \mathrm{~F}(x) \mathrm{d} x$ is equal to :

Aicharya · Accepted Answer

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

Let $\mathrm{f}: \mathbf{R} \rightarrow \mathbf{R}$ be a twice differentiable function such that $f(2)=1$.…

Solution