JEE Main 2021 — Mathematics Calculus

Let $\displaystyle I={\int }_{-\frac{1}{2}}^{1}([2x]+|x|)dx$I=∫−21​1​([2x]+∣x∣)dx

$[2x]$[2x] will be discontinuous at $x=0,\frac{1}{2};x\in (\frac{-1}{2},1)$x=0,21​;x∈(2−1​,1)

We know that $|x|=x,x\geq 0;-x,x<0.$∣x∣=x,x≥0;−x,x<0.

$$={\int }_{-\frac{1}{2}}^{0}(-1-x)dx+{\int }_{0}^{1/2}(0+x)dx+{\int }_{\frac{1}{2}}^{1}(1+x)dx$$=∫−21​0​(−1−x)dx+∫01/2​(0+x)dx+∫21​1​(1+x)dx

We know that $\displaystyle \int {x}^{n}dx={x}^{\frac{n+1}{n+1}}+c$∫xndx=xn+1n+1​+c

$$={(-x-\frac{{x}^{2}}{2})}_{-\frac{1}{2}}^{0}+{(\frac{{x}^{2}}{2})}_{0}^{1/2}+{(x+\frac{{x}^{2}}{2})}_{\frac{1}{2}}^{1}$$=(−x−2x2​)−21​0​+(2x2​)01/2​+(x+2x2​)21​1​

$$=(0)-(+\frac{1}{2}-\frac{1}{8})+\frac{1}{8}+(1+\frac{1}{2})-(\frac{1}{2}+\frac{1}{8})=\frac{5}{8}$$=(0)−(+21​−81​)+81​+(1+21​)−(21​+81​)=85​

$\displaystyle \Rightarrow 8{\int }_{-\frac{1}{2}}^{1}([2x]+|x|)dx=5$⇒8∫−21​1​([2x]+∣x∣)dx=5.