Let be a differentiable function such that for all . Then the area of the region bounded by y = f(x) and the coordinates axes is [2025]

Question

Let $f : [0, \infty) \to R$ be a differentiable function such that $f (x) = 1 - 2 x + \int_{0}^{x} e^{x - t} f (t) d t$ for all $x \in [0, \infty)$ .

Then the area of the region bounded by y = f(x) and the coordinates axes is [2025]

Smart Achievers · Accepted Answer

(1)

We have, $y = 1 - 2 x + e^{x} \int_{0}^{x} e^{- t} f (t) d t$

$\Rightarrow \frac{d y}{d x} = - 2 + e^{- x} \cdot e^{x} f (x) + e^{x} \int_{0}^{x} e^{- t} f (t) d t$

$= - 2 + y + y + 2 x - 1$

$\Rightarrow \frac{d y}{d x} - 2 y = 2 x - 3$

This is a linear differential equation, so we have

$y e^{- 2 x} = \int (2 x - 3) \cdot e^{- 2 x} d x + c$ [ $∵ I . F . = e^{\int - 2 d x} = e^{- 2 x}$ ]

$\Rightarrow y e^{- 2 x} = - \frac{(2 x - 3)}{2} e^{- 2 x} + \int e^{- 2 x} \cdot d x + c$

$\Rightarrow y e^{- 2 x} = \frac{- (2 x - 3)}{2} e^{- 2 x} - \frac{e^{- 2 x}}{2} + c$

$∵ f (0) = 1 ∴ c = 1 - \frac{3}{2} + \frac{1}{2} = 0$

$\Rightarrow y = - \frac{(2 x - 3)}{2} - \frac{1}{2} \Rightarrow x + y = 1$

So, area $= \frac{1}{2} \times 1 \times 1 = \frac{1}{2}$ .

Want Us to Email you About Special Offers & Updates?