JEE Main 2023 — Mathematics Calculus

Given,

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

\Rightarrow f(x)={\begin{matrix}{e}^{{x}^{2}} & x\in [0,1) \\ e, & x\in [1,2]\end{matrix}

As $x-\mathrm{ln}x\in [1,2)$x−lnx∈[1,2) for $x\in [1,2]$x∈[1,2], so$[x-\mathrm{ln}x]=1$[x−lnx]=1 and $x-[x]={x}$x−[x]=x, so $x\in (0,1),{x}^{2}<{x}$x∈(0,1),x2<x

Now solving the integral we get,$\displaystyle {\int }_{0}^{2}xf(x)={\int }_{0}^{1}x\cdot {e}^{{x}^{2}}dx+{\int }_{1}^{2}x\cdot edx$∫02​xf(x)=∫01​x⋅ex2dx+∫12​x⋅edx

Now in first integral, let ${x}^{2}=t\Rightarrow 2xdx=dt$x2=t⇒2xdx=dt we get,

$$\Rightarrow {\int }_{0}^{2}xf(x)=[\frac{1}{2}{\int }_{0}^{1}{e}^{t}dt]+{[e\cdot \frac{{x}^{2}}{2}]}_{1}^{2}$$⇒∫02​xf(x)=[21​∫01​etdt]+[e⋅2x2​]12​

$$\Rightarrow {\int }_{0}^{2}xf(x)=(\frac{e-1}{2})+\frac{3e}{2}$$⇒∫02​xf(x)=(2e−1​)+23e​

$$\Rightarrow {\int }_{0}^{2}xf(x)=2e-\frac{1}{2}$$⇒∫02​xf(x)=2e−21​