Application of IntegralsmediumPYQ · JEE Main · 4 Apr 2026 · Shift 1 (Morning)Free

Area Bounded by π-|x| and |x sin x|: 2 + π²/4 | JEE 2026

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

Question
The area of the region {(x,y):y≤π−∣x∣, y≤∣xsin⁡x∣, y≥0}\{(x,y):y\le\pi-|x|,\ y\le|x\sin x|,\ y\ge0\} is
A1+π281+\dfrac{\pi^2}{8}
B2+π242+\dfrac{\pi^2}{4}correct
Cπ28−1\dfrac{\pi^2}{8}-1
D4+π224+\dfrac{\pi^2}{2}
Solution
Step 1: By symmetry (even functions), area =2×=2\times (area for x≥0x\ge0). For 0≤x≤π0\le x\le\pi, the boundary ∣xsin⁡x∣=xsin⁡x|x\sin x|=x\sin x; the relevant curve switches with π−x\pi-x at x=π/2x=\pi/2. Step 2:
Area=2(∫0π/2xsin⁡x dx+∫π/2π(π−x) dx).\text{Area}=2\left(\int_0^{\pi/2}x\sin x\,dx+\int_{\pi/2}^{\pi}(\pi-x)\,dx\right).
 Step 3: Evaluate:
∫0π/2xsin⁡x dx=[−xcos⁡x+sin⁡x]0π/2=1,∫π/2π(π−x) dx=[(π−x)2−2]...=π28.\int_0^{\pi/2}x\sin x\,dx=\big[-x\cos x+\sin x\big]_0^{\pi/2}=1,\qquad \int_{\pi/2}^{\pi}(\pi-x)\,dx=\left[\frac{(\pi-x)^2}{-2}\right]... =\frac{\pi^2}{8}.
 Step 4:
Area=2(1+π28)=2+π24.\text{Area}=2\left(1+\frac{\pi^2}{8}\right)=2+\frac{\pi^2}{4}.
 Correct answer: (2)
Solution working
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