<p>Area of the region \(y^2\le2x\) and \(y\ge x-4\). [JEE Main 2020]</p>
Step-by-Step Solution
Key Concept: Parabola x=y^2/2 and line x=y+4. Intersections: y^2/2=y+4 \to y^2-2y-8=0 \to y=4,-2. Integrate w.r.t. y.
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<p>Intersections: $y^2=2(y+4)\Rightarrow y^2-2y-8=(y-4)(y+2)=0\Rightarrow y=-2,4$.</p>
<p>On $[-2,4]$: line $x=y+4\ge$ parabola $x=y^2/2$.</p>
<p>$$A=\int_{-2}^4\left[(y+4)-\frac{y^2}{2}\right]dy=\left[\frac{y^2}{2}+4y-\frac{y^3}{6}\right]_{-2}^4$$</p>
<p>$=(8+16-\frac{64}{6})-(2-8+\frac{8}{6})=24-\frac{64}{6}-(-6+\frac{8}{6})=30-\frac{72}{6}=30-12=18$. ✓</p>
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Correct Answer: A