Question

Let $f: [0, \infty) \to \mathbb{R}$ be a differentiable function such that $f(x) = 1 - 2x + \int_0^x e^{x-t} f(t) \, dt$ for all $x \in [0, \infty)$. Then the area of the region bounded by $y = f(x)$ and the coordinate axes is

• A. $\sqrt{5}$
• B. $\frac{1}{2}$
• C. $\sqrt{2}$
• D. $2$

Hint:

Differentiate both sides of the integral equation to simplify.

Answer 1

The Correct Option is B

1. Differentiate $f(x)$: \[ f'(x) = -2 + e^x \int_0^x e^{-t} f(t) \, dt + e^x e^{-x} f(x) \] \[ f'(x) = -2 + e^x \int_0^x e^{-t} f(t) \, dt + f(x) \]
2. Simplify the differential equation: \[ f'(x) - f(x) = -2 \]
3. Solve the differential equation: \[ \frac{d}{dx} \left( e^{-x} f(x) \right) = -2e^{-x} \] \[ e^{-x} f(x) = \int -2e^{-x} \, dx = 2e^{-x} + c \] \[ f(x) = 2 + ce^x \]
4. Use the initial condition $f(0) = 1$: \[ 1 = 2 + c \implies c = -1 \] \[ f(x) = 2 - e^x \] 5. Find the area under the curve $y = f(x)$: \[ \text{Area} = \int_0^\infty (2 - e^x) \, dx \] \[ \text{Area} = \left[ 2x - e^x \right]_0^\infty = \left[ 2x - e^x \right]_0^\infty = \frac{1}{2} \] Therefore, the correct answer is (2) $\frac{1}{2}$.