Subject: Mathematics — Definite Integration
Question: Let $f:\lbrack 0,\infty) \rightarrow \mathbb{R}$ be a
continuous function such that
$f(x) = 1 - 2x + \int_{0}^{x}\mspace{2mu} e^{x - t}f(t)dt$\
for all $x \in \lbrack 0,\infty)$. Then, find the which of the area of
the region
$\left\{ (x,y) \in \lbrack 0,1\rbrack \times \mathbb{R:}f(x) \leq y \leq \sqrt{1 - x^{2}} \right\}$
Answer: $\frac{\pi - 2}{4}$
Solution: $f(x) = 1 - 2x + \int_{0}^{x}\mspace{2mu} e^{x - t}f(t)dt$\
$${\Rightarrow \ f(x) = 1 - 2x + e^{x}\int_{0}^{x}\mspace{2mu}\mspace{2mu} e^{- t}f(t)dt
}{\Rightarrow \ f^{'}(x) = - 2 + e^{x}\int_{0}^{x}\mspace{2mu}\mspace{2mu} e^{- t}f(t)dt + e^{x}\left\lbrack e^{- x}f(x) \right\rbrack
}{\Rightarrow \ f^{'}(x) = - 2 + \lbrack f(x) - 1 + 2x\rbrack + f(x)
}{\Rightarrow \ f^{'}(x) - 2f(x) = 2x - 3
}$$Its a linear differential equation.\
$IF = e^{\int_{}^{}\ - 2dx} = e^{- 2x}$\
Solution: $f(x) \times e^{- 2x} = \int_{}^{}\ e^{- 2x}(2x - 3)dx$\
$$\Rightarrow f(x) \times e^{- 2x} = \frac{e^{- 2x}}{- 2}(2x - 3) - \int_{}^{}\ \frac{e^{- 2x}}{- 2} \times 2dx
$$From definition of function, $f(0) = 1$\
$$\therefore 1 = 1 + c \Rightarrow c = 0,\therefore f(x) = 1 - x
$$Clearly curve $y = 1 - x$, does not pass through $(1,2)$ but it passes
through $(2, - 1)$
[IMAGE] \
∴ (a) is false and (b) is true. Also the area of the region
$1 - x \leq y \leq \sqrt{1 - x^{2}}$, is shown in the figure by the
shaded region, is given by = Area of quadrant - Area
$\bigtriangleup OAB$\
$$= \frac{1}{4} \times \pi \times 1^{2} - \frac{1}{2} \times 1 \times 1 = \frac{\pi - 2}{4}$$