Let f:[0, ) R be a differentiable function such that f(x)=1-2 x+_0^x e^x-t f(t) d t for al…

Question

Let $f:[0, \infty) \rightarrow \mathbb{R}$ 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 coordinate axes is

Accepted Answer

$$\because f(x)=1-2 x+\int_0^x e^{x-t} f(t) d t$$

or, $f(x)=1-2 x+e^x \int_0^x e^{-t} f(t) d t$

on differentiating both sides w.r.t. $x$ we get

$$\begin{aligned}
& f^{\prime}(x)=-2+e^x \int_0^x e^{-t} f(t) d t+e^x \cdot e^{-x} f(x) \
& f^{\prime}(x)=-2+f(x)+2 x-1+f(x)\{\text { from eq. (1) \}}
\end{aligned}$$

$$\begin{aligned}
& \therefore \quad f(x)-2 f(x)=2 x-3 \
& \text { I.F. }=e^{\int-2 d x}=e^{-2 x} \
& \therefore \quad e^{-2 x} \cdot f(x)=\int e^{-2 x}(2 x-3) d x \
& e^{-2 x} \cdot f(x)=(2 x-3) \cdot \frac{e^{-2 x}}{-2}-\int 2 \cdot \frac{e^{-2 x}}{-2} d x \
& e^{-2 x} \cdot f(x)=\frac{(2 x-3) e^{-2 x}}{-2}+\frac{e^{-2 x}}{-2}+c \
& f(x)=-x+1+c^{\prime} e^{2 x} \
& \because \quad f(x)=1 \text { from eq. (1) } \
& \therefore \quad 1=0+1+c^{\prime} \Rightarrow c^{\prime}=0 \
& \therefore \quad f(x)=-x+1
\end{aligned}$$

$\Rightarrow \quad$ Area $=\frac{1}{2}$

Solutions · 1