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∫ max(x+|x|, x−[x]) dx from −2 to 2 | JEE

JEE Maths question with a full step-by-step solution.

Question

\displaystyle\int_{-2}^{2}\max\big(x+|x|,\ x-[x]\big)\,dx

, where

[x]

denotes the greatest integer

\le x

, equals

4

5

correct

6

8

Solution

Step 1: Simplify each candidate. We have

x+|x|=\begin{cases}2x,&x\ge 0\\[2pt]0,&x<0\end{cases}

, and

x-[x]=\{x\}

, the fractional part, which always lies in

[0,1)

. Step 2: Determine the maximum on each side. For

x\ge 0

2x\ge x\ge\{x\}

, so the maximum is

2x

. For

x<0

x+|x|=0

while

\{x\}\ge 0

, so the maximum is

\{x\}

. Hence

\max\big(x+|x|,\ x-[x]\big)=\begin{cases}2x,&x\ge 0\\[2pt]\{x\},&x<0.\end{cases}

Step 3: Integrate the

x\ge 0

part:

\int_{0}^{2}2x\,dx=\big[x^{2}\big]_{0}^{2}=4.

Step 4: Integrate the

x<0

part using

\{x\}=x+2

[-2,-1)

and

\{x\}=x+1

[-1,0)

\int_{-2}^{-1}(x+2)\,dx+\int_{-1}^{0}(x+1)\,dx=\left[\dfrac{x^{2}}{2}+2x\right]_{-2}^{-1}+\left[\dfrac{x^{2}}{2}+x\right]_{-1}^{0}=\dfrac12+\dfrac12=1.

Adding the two parts,

\displaystyle\int_{-2}^{2}\max(\cdots)\,dx=4+1=5.

Correct answer: (2)

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