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 ∫₀² f(x) dx is :

Solution

Maths sir · 4 steps

The derivative condition says f(x) + f(2-x) never changes, so pair the integral with its mirror image.

1. Build a constant. By the chain rule, d/dxf(2-x) = -f'(2-x). So d/dx[f(x) + f(2-x)] = f'(x) - f'(2-x) = 0, and f(x) + f(2-x) is a constant on [0, 2].
2. Find the constant. Put x = 0: f(0) + f(2) = 1 + e². So f(x) + f(2-x) = 1 + e² for every x in [0, 2].
3. Use the mirror property. Let I = ∫₀² f(x) dx. Replacing x by 2 - x gives I = ∫₀² f(2-x) dx. Add the two forms of I.

   2I = ∫₀² [f(x) + f(2-x)]dx = ∫₀² (1 + e²) dx = 2(1 + e²)

4. So. I = 1 + e², which is option (a).

The idea to keep. When you are told f'(x) = f'(a-x), it means f(x) + f(a-x) is constant. Add ∫₀ᵃ f(x) dx to its mirror ∫₀ᵃ f(a-x) dx and the integral falls out.

Answer from the NTA final answer key. Solution by PrepWiser.