Question

Let f(x) be a differentiable function defined on [0, 2] such that f'(x) = f'(2-x) for all x ∈ (0, 2), f(0) = 1 and f(2) = e². Then the value of ∫ from 0 to 2 f(x) dx is :

• A. 1 - e²
• B. 1 + e²
• C. 2(1 + e²)
• D. 2(1 - e²)

Hint:

King's Property: $\int_a^b f(x)dx = \int_a^b f(a+b-x)dx$. Adding the two integrals often simplifies the integrand to a constant.

Answer 1

The Correct Option is B

Step 1: Integrate $f'(x) = f'(2-x)$. We get $f(x) = -f(2-x) + C$.
Step 2: Find $C$ using $x=0$: $f(0) = -f(2) + C \Rightarrow 1 = -e^2 + C \Rightarrow C = 1 + e^2$.
Step 3: Thus, $f(x) + f(2-x) = 1 + e^2$.
Step 4: Use the property $\int_0^a f(x)dx = \int_0^a f(a-x)dx$.
Step 5: $2I = \int_0^2 [f(x) + f(2-x)]dx = \int_0^2 (1+e^2)dx = (1+e^2)[x]_0^2 = 2(1+e^2)$.
Step 6: $I = 1+e^2$.