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Let f(x)= … . and h(x)=f(|x|)+|f(x)|. Then ∫_-2^2 h(x) d x is equal to :

JEE Main 2024 — Mathematics Calculus

2024mcqmedium

Let f(x)={−2,−2≤x≤0x−2,0<x≤2f(x)=\left\{\begin{array}{lr}-2, & -2 \leq x \leq 0 \\ x-2, & 0 < x \leq 2\end{array}\right.f(x)={−2,x−2,​−2≤x≤00<x≤2​ and h(x)=f(∣x∣)+∣f(x)∣h(x)=f(|x|)+|f(x)|h(x)=f(∣x∣)+∣f(x)∣. Then ∫−22h(x)dx\displaystyle \int_{-2}^2 h(x) \mathrm{d} x∫−22​h(x)dx is equal to :

Official previous-year question

Held on 4 Apr 2024 · Verified 6 Jul 2026.

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