JEE Main 2023 (Online) 11th April Evening ShiftQuestion 1
Let the function $f:[0,2]\to\mathbb{R}$ be defined as $f(x)=\begin{cases}e^{\min\{x^{2},x-[x]\}}, & x\in[0,1)\\ e^{[x-\log_ex]}, & x\in[1,2]\end{cases}$ where $[t]$ denotes the greatest integer less than or equal to $t$. Then the value of the integral $\int_{0}^{2}xf(x)\,dx$ is
Definite Integration
Solution
Concept. Simplify $f$ on $[0,1)$ and on $[1,2]$ separately; both pieces become easy. Formulas used. (1) For $0\le x<1$: $x-[x]=x$ and $x^2\le x$(2) $x-\ln x$ is increasing for $x\ge1$ (derivative $1-\frac1x\ge0$)Step 1 - $[0,1)$. By (1), $\min\{x^2,x\}=x^2$, so $f=e^{x^2}$: $$\int_0^1xe^{x^2}dx=\frac12\big[e^{x^2}\big]_0^1=\frac{e-1}2.$$ Step 2 - $[1,2]$. By (2), $x-\ln x$ goes from $1$ to $2-\ln2\approx1.31$, so $[x-\ln x]=1$ and $f=e$: $$\int_1^2ex\,dx=\frac{3e}2.$$ Step 3 - Add. $\frac{e-1}2+\frac{3e}2=2e-\frac12$. Answer: (C)