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

(D) $2\left(1-\mathrm{e}^{2}\right)$

JEE Main · 2021 · 24 Feb 2021 · Shift 2 · Definite Integrals

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) = e^{2}

. Then the value of

\int_{0}^{2} f (x) d x

Answer: (C)

Put

g (x) = f (x) + f (2 - x)

. Then

g^{'} (x) = f^{'} (x) - f^{'} (2 - x) = 0

(0, 2)

, so

g

is constant and

g (x) = g (0) = f (0) + f (2) = 1 + e^{2}

. The substitution

x \mapsto 2 - x

gives

\int_{0}^{2} f (x) d x = \int_{0}^{2} f (2 - x) d x

, so

2 \int_{0}^{2} f (x) d x = \int_{0}^{2} [f (x) + f (2 - x)] d x = \int_{0}^{2} (1 + e^{2}) d x = 2 (1 + e^{2}) .

Dividing by

2

\int_{0}^{2} f (x) d x = 1 + e^{2}

JEE Main · 2021 · 24 Feb 2021 · Shift 2 · Single correct option