JEE Main 2024 — Definite Integration Question with Solution
From: JEE Main 2024 (Online) 4th April Morning Shift
Question
\text { Let } f(x)=\left\{\begin{array}{lr} -2, & -2 \leq x \leq 0 \\ x-2, & 0< x \leq 2 \end{array} \text { and } \mathrm{h}(x)=f(|x|)+|f(x)| \text {. Then } \int_\limits{-2}^2 \mathrm{~h}(x) \mathrm{d} x\right. \text { is equal to: }
Choose an option
Show full solutionCorrect option: A
Step-by-step explanation
\begin{aligned} & h(x)=\left\{\begin{array}{cc} -x-2+2=-x & -2 \leq x \leq 0 \\ 0 & 0< x \leq 2 \end{array}\right. \\ & \therefore \int_\limits{-2}^2 h(x) d x=\int_\limits{-2}^0-x d x+\int_\limits0^2 0 d x \\ & \left.\frac{x^2}{2}\right|_{-2} ^0=\frac{4}{2}=2 \end{aligned}
Practice this on the real CBT interface
Solve this JEE Main question (and the rest of the Definite Integration chapter) on PrepSharp's TCS iON-style CBT player — with timer, bookmarks and session analytics.
Solve interactively →About this question
This is a previous-year question from JEE Main 2024, covering the Definite Integration chapter of Mathematics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.