Let $f:[0, \infty) \rightarrow \mathbb{R}$ be differentiable function such that $f(\mathrm{x})=1-2…
Let $f:[0, \infty) \rightarrow \mathbb{R}$ be differentiable function such that $f(\mathrm{x})=1-2 \mathrm{x}+\int_0^x e^{x-t} f(t) \mathrm{dt}$ for all $\mathrm{x} \in[0, \infty)$. Then the area of the region bounded by $\mathrm{y}=f(\mathrm{x})$ and the coordinate axes is
$\sqrt{5}$
$\frac{1}{2}$
$\sqrt{2}$
2
Solution
$\begin{aligned} & y=1-2 x+e^x \int_0^x e^{-t} f(t) d t \\ & \frac{d y}{d x}=-2+e^{-x} \cdot e^x f(x)+e^x \int_0^x e^{-t} f(t) d t \\ & \frac{d y}{d x}=-2+y+y+2 x-1 \\ & \frac{d y}{d x}-2 y=(2 x-3) \\ & y e^{-2 x}=\int(2 x-3) d x \cdot e^{-2 x} \\ & y e^{-2 x}=\frac{-(2 x-3)}{2} e^{-2 x}+\int e^{-2 x} d x \\ & y e^{-2 x}=\frac{-(2 x-3)}{2} e^{-2 x}-\frac{1}{2} e^{-2 x}+c \\ & f(0)=1 \Rightarrow c=1-\frac{3}{2}+\frac{1}{2}=0\end{aligned}$ $\begin{aligned} & y=-\frac{(2 x-3)}{2}-\frac{1}{2} \\ & y=-x+1 \\ & x+y=1 \\ & \text { area }=\frac{1}{2}(1)(1)=\frac{1}{2}\end{aligned}$