Let the function [f:[0,2] R] be defined as [f(x)= casese^ x^2, x-[x], & x [0,1) e^[x- _e x…

Question

Let the function $$f:[0,2] \rightarrow \mathbb{R}$$ be defined as

$$f(x)= \begin{cases}e^{\min \left\{x^{2}, x-[x]\right\},} & x \in[0,1) \ e^{\left[x-\log _{e} x\right]}, & x \in[1,2]\end{cases}$$

where $$[t]$$ denotes the greatest integer less than or equal to $t$. Then the value of the integral $$\int_\limits{0}^{2} x f(x) d x$$ is :

Accepted Answer

$$
\begin{aligned}
\operatorname{Minimum}\left\{\mathrm{x}^2,\{\mathrm{x}\}\right\} & =\mathrm{x}^2 ; \mathrm{x} \in[0,1) \\
{\left[\mathrm{x}-\log _{\mathrm{e}} \mathrm{x}\right] } & =1 ; \mathrm{x} \in[1,2)
\end{aligned}
$$

$$
\therefore \mathrm{f}(\mathrm{x})=\left\{\[\begin{gathered} \mathrm{e}^{\mathrm{x}^2} ; \mathrm{x} \in[0,1) \ \mathrm{e} ; \mathrm{x} \in[1,2) \end{gathered}$$\right.
\]

$$
\begin{aligned}
& \int\limits_0^2 x f(x)=\int\limits_0^1 x \cdot e^{x^2} d x+\int\limits_1^2 x \cdot e d x \\
& x^2=t \Rightarrow 2 x d x=d t
\end{aligned}
$$

$$
=\frac{1}{2} \int\limits_0^1 e^t d t+\left.e \frac{x^2}{2}\right|_1 ^2
$$

$$
\begin{aligned}
& =\frac{1}{2}(\mathrm{e}-1)+\frac{1}{2}(4-1) \mathrm{e} \\
& =2 \mathrm{e}-\frac{1}{2}
\end{aligned}
$$

Solutions · 1