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