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Answer: \(8\sqrt{3}\) sq. units

The parabola \(y^2=4x\) is symmetric about the X-axis. The required region lies between \(x=0\) and \(x=3\), so

Required area \(=2\times\) area above the X-axis \(=2\int_0^3 y\,dx=2\int_0^3 2\sqrt{x}\,dx\)
\(=4\left[\frac{x^{3/2}}{3/2}\right]_0^3=\frac{8}{3}\left[3^{3/2}-0\right]=\frac{8}{3}\times 3\sqrt{3}=8\sqrt{3}\) sq. units.

Doubtly AI · 29 Oct 2026 AI-verified · Nov 2026 Report
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