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
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
Show Hint
- $\sqrt{5}$
- $\frac{1}{2}$
- $\sqrt{2}$
- $2$
The Correct Option is B
Approach Solution - 1
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) $\frac{1}{2}$.
Approach Solution -2
We are given a differentiable function \( f:[0,\infty)\to\mathbb{R} \) defined by \[ f(x)=1-2x+\int_0^x e^{\,x-t}f(t)\,dt, \] and we are asked to find the area of the region bounded by \( y=f(x) \) and the coordinate axes.
Concept Used:
Convert the given Volterra-type integral equation into a first-order linear ODE by differentiating. Use the initial value obtained from the integral equation at \( x=0 \) to determine the unique solution. The area bounded by the curve and coordinate axes (in the first quadrant) is the area of the triangle formed by the intercepts.
Step-by-Step Solution:
Step 1: Rewrite the integral term and differentiate \( f \).
\[ \int_0^x e^{x-t}f(t)\,dt = e^{x}\int_0^x e^{-t}f(t)\,dt = e^{x}g(x),\quad\text{where } g(x):=\int_0^x e^{-t}f(t)\,dt. \] \[ \Rightarrow\ f(x)=1-2x+e^{x}g(x). \] Differentiate both sides: \[ f'(x)=-2+\frac{d}{dx}\big(e^{x}g(x)\big)=-2+e^{x}g(x)+e^{x}g'(x). \] Since \( g'(x)=e^{-x}f(x) \), we get \[ f'(x)=-2+e^{x}g(x)+f(x). \]Step 2: Eliminate \( g \) using the defining relation for \( f \).
From \( f(x)=1-2x+e^{x}g(x) \), we have \( e^{x}g(x)=f(x)-1+2x \). Substitute: \[ f'(x)=-2+\big(f(x)-1+2x\big)+f(x)= -3+2x+2f(x). \] Thus, \[ f'(x)-2f(x)=2x-3. \]Step 3: Determine the initial condition from the original equation at \( x=0 \).
\[ f(0)=1-0+\int_0^0(\cdots)\,dt=1. \]Step 4: Solve the linear ODE \( f'-2f=2x-3 \) with \( f(0)=1 \).
Using the integrating factor \( e^{-2x} \): \[ \frac{d}{dx}\big(f(x)e^{-2x}\big)= (2x-3)e^{-2x}. \] Integrate: \[ f(x)e^{-2x}= \int (2x-3)e^{-2x}\,dx + C. \] A direct computation gives \[ \int (2x-3)e^{-2x}\,dx = e^{-2x}(1-x)+C, \] hence \[ f(x)e^{-2x}= e^{-2x}(1-x)+C \ \Rightarrow\ f(x)=1-x+Ce^{2x}. \] Use \( f(0)=1 \) to find \( C=0 \). Therefore, \[ f(x)=1-x\quad (x\ge 0). \]Step 5: Compute the required area bounded by \( y=f(x)=1-x \) and the coordinate axes.
\[ \text{Intercepts: } y\text{-intercept }(0,1),\ \ x\text{-intercept }(1,0). \] This forms a right triangle in the first quadrant with base \(1\) and height \(1\): \[ \text{Area}=\frac{1}{2}\times 1\times 1=\frac{1}{2}. \]Final Computation & Result
The area of the region bounded by \( y=f(x) \) and the coordinate axes is \(\dfrac{1}{2}\).
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