Let f:[0,∞)→ ℝ be a differentiable function such that
f(x)=1-2x+∫₀ˣeˣ⁻ᵗf(t)dt for all x∈ [0,∞).
Then the area of the region bounded by y=f(x) and the coordinate axes is

Solution

Maths sir · 4 steps

Differentiate the integral equation once and it becomes a linear ODE whose solution is a line.

1. Find f(0). Put x = 0. The integral vanishes, so f(0) = 1.
2. Differentiate. By the Leibniz rule, d/dx∫₀ˣ eˣ⁻ᵗf(t) dt = f(x) + ∫₀ˣ eˣ⁻ᵗf(t) dt. From the given equation, that integral equals f(x) - 1 + 2x, so f'(x) = -2 + f(x) + f(x) - 1 + 2x.

   f'(x) - 2f(x) = 2x - 3

3. Solve the ODE. Try f = ax + b: a - 2ax - 2b = 2x - 3 gives a = -1 and b = 1. So f(x) = 1 - x + Ce²ˣ, and f(0) = 1 forces C = 0. That leaves f(x) = 1 - x.
4. So. The line y = 1 - x meets the axes at (1, 0) and (0, 1). The enclosed triangle has area 1/2 × 1 × 1 = 1/2, which is option (d).

The idea to keep. When x sits both outside and inside an integral equation, differentiate it and replace the integral using the original equation. The initial value comes from putting x = 0.

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