Let f: ℝ → ℝ be a differentiable function such that f′(x)+f(x)=∫0² f(t) d t. If f(0)=e⁻², then 2 f(0)-f(2) is equal to ____________.

Solution

Maths sir · 4 steps

That integral is just a constant, so call it k and solve a very easy linear equation.

1. Name the constant. ∫₀² f(t) dt is a fixed number. Call it k, so the equation becomes f'(x) + f(x) = k.
2. Solve the equation. The integrating factor is eˣ, so (eˣ f)' = k eˣ. Integrating gives f(x) = k + C e⁻ˣ.
3. Find k and C. Put this f back into the definition of k: k = ∫₀² (k + C e⁻ᵗ) dt = 2k + C(1 - e⁻²), so k = -C(1 - e⁻²). Then f(0) = k + C = C e⁻², and f(0) = e⁻² gives C = 1 and k = e⁻² - 1.

   f(x) = e⁻ˣ + e⁻² - 1

4. So. f(2) = 2e⁻² - 1, so 2f(0) - f(2) = 2e⁻² - (2e⁻² - 1) = 1.

The idea to keep. A definite integral with fixed limits is only a number. Replace it with a constant, solve the equation, then substitute back to find that constant.

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