Let f(x) be a differentiable function
Question:

Let $f(x)$ be a differentiable function defined on $[0,2]$ such that $f^{\prime}(x)=f^{\prime}(2-x)$ for all $x \in(0,2), f(0)=1$ and $f(2)=\mathrm{e}^{2}$. Then the value of $\int_{0}^{2} f(x) \mathrm{d} x$ is:

1. (1) $1+e^{2}$

2. (2) $1-\mathrm{e}^{2}$

3. (3) $2\left(1-e^{2}\right)$

4. (4) $2\left(1+e^{2}\right)$

Correct Option: 1

Solution:

$f^{\prime}(x)=f^{\prime}(2-x)$

On integrating both side $f(x)=-f(2-x)+c$

put $x=0 f(0)+f(2)=c \quad \Rightarrow c=1+e^{2}$

$\Rightarrow \mathrm{f}(\mathrm{x})+\mathrm{f}(2-\mathrm{x})=1+\mathrm{e}^{2} \ldots \ldots$ (i)

$I=\int_{0}^{2} f(x) d x=\int_{0}^{1}\{f(x)+f(2-x)\} d x=\left(1+e^{2}\right)$