Let the function $f:\left[0,2\right]\to \mathrm{ℝ}$ be defined as $f\left(x\right)=\left\{\begin{array}{cc}{e}^{\min \left\{x^2,x-\left[x\right]\right\}},& x\in [0,1)\\ e^{\left[x-{\log }_{e}x\right]},& x\in \left[1,2\right]\end{array}\right.$, where $\left[t\right]$ denotes the greatest integer less than or equal to $t$. Then the value of the integral ${\int }_0^{2}xf\left(x\right)dx$ is

Given,

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

As $x-\ln x\in [1,2)$ for $x\in [1,2]$, so$\left[x-\ln x\right]=1$ and $x-\left[x\right]=\left\{x\right\}$, so $x\in \left(0,1\right), x^2<\left\{x\right\}$

Now solving the integral we get,${\int }_0^{2}xf\left(x\right)={\int }_0^{1}x·e^{x^2}dx+{\int }_1^{2}x·e dx$

Now in first integral, let $x^2=t\Rightarrow 2xdx=dt$ we get,

$\Rightarrow {\int }_0^{2}xf\left(x\right)=\left[\frac{1}{2}{\int }_0^1{e}^t dt\right]+{\left[e·\frac{x^2}{2}\right]}_1^2$

$\Rightarrow {\int }_0^{2}xf\left(x\right)=\left(\frac{e-1}{2}\right)+\frac{3e}{2}$

$\Rightarrow {\int }_0^{2}xf\left(x\right)=2e-\frac{1}{2}$