Question:
Let $\mathrm{f}(\mathrm{x})$ 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 $f(2)=e^{2}$. Then the value of $\int_{0}^{2} f(x) d x$
Correct Option: , 2
Solution:
$f^{\prime}(x)=f^{\prime}(2-x)$
$f(x)=-f(2-x)+c$
put $x=0$
$\mathrm{f}^{\prime}(0)=-\mathrm{f}^{\prime}(2)+\mathrm{c}$
$\mathrm{c}=\mathrm{f}(0)+\mathrm{f}(2)=1+\mathrm{e}^{2}$
so, $\mathrm{f}(\mathrm{x})+\mathrm{f}(2-\mathrm{x})=1+\mathrm{e}^{2}$
$I=\int_{0}^{2} f(x) d x$
$I=\int_{0}^{2} f(2-x) d x$
$2 I=\int_{0}^{2}(f(x)+f(2-x)) d x$
$2 I=\left(1+e^{2}\right) \int_{0}^{2} d x$
$I=1+e^{2}$
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