"Let f(x) be a differentiable function defined on [0, 2] such that f'(x) = f'(2 $-$ x) for all x$\in$ (0, 2), f(0) = 1 and f(2) = e2. Then the value of $\int\limits_0^2 {f(x)} dx$ is :"

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Accepted Answer

"f'(x) = f'(2 $-$ x)

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

put x = 0

f(0) + f(2) = c $\Rightarrow$ c = 1 + e²

$\Rightarrow$ f(x) + f(2 $-$ x) = 1 + e² ..... (i)

$$I = \int\limits_0^2 {f(x)dx} = \int\limits_0^1 {\{ f(x) + f(2 - x)\} dx = (1 + {e^2})} $$"

Let f(x) be a differentiable function defined on [0, 2] such that f'(x) = f'(2 $-$ x) for all x$\in$ (0, 2), f(0) = 1 and f(2) = e². Then the value of $\int\limits_0^2 {f(x)} dx$ is :

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Let f(x) be a differentiable function defined on [0, 2] such that f'(x) = f'(2 $-$ x) for all x$\in$ (0, 2), f(0) = 1 and f(2) = e2. Then the value of $\int\limits_0^2 {f(x)} dx$ is :

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Let f(x) be a differentiable function defined on [0, 2] such that f'(x) = f'(2 $-$ x) for all x$\in$ (0, 2), f(0) = 1 and f(2) = e². Then the value of $\int\limits_0^2 {f(x)} dx$ is :