Let the function f: [0, 2] → ℝ be defined as f(x) = \ cases e ^...

Given,

f\left(x\right) = \left\{\begin{cases}{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{cases}\right)<br />⇒ f\left(x\right) = \left\{$\begin{cases}{e}^{{x}^{2}} & x \in [ 0, 1 ) \\ e, & x \in \left[1, 2\right]\end{cases}$\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,${∫}_{0}^{2}xf\left(x\right) = {∫}_{0}^{1}x \cdot {e}^{{x}^{2}}dx + {∫}_{1}^{2}x \cdot edx$

Now in first integral, let ${x}^{2} = t ⇒ 2xdx = dt$ we get,

$⇒ {∫}_{0}^{2}xf\left(x\right) = \left[\dfrac{1}{2}{∫}_{0}^{1}{e}^{t}dt\right] + {\left[e \cdot \dfrac{{x}^{2}}{2}\right]}_{1}^{2}$

$⇒ {∫}_{0}^{2}xf\left(x\right) = \left(\dfrac{e - 1}{2}\right) + \dfrac{3e}{2}$

$⇒ {∫}_{0}^{2}xf\left(x\right) = 2e - \dfrac{1}{2}$