Q.3 (A)(iii) Find the area of region bounded by y² = 25x and the line x = 4.

Answer: 160/3 sq. units y² = 25x is a parabola with vertex at the origin, symmetric about the X-axis. The line x = 4 meets it at y = ± 10, i.e. at (4, 10) and (4, -10). By symmetry, required area = 2 × (area above the X-axis), where y = 5√(x): displaystyle A = 2∫₀⁴ 5√(x) dx = 10[fracx^(3/2)3/2]₀⁴ = 10 · 2/3(4^(3/2) - 0) = 20/3 × 8 = 160/3 sq. units.

Chapter: Application of Definite Integration· Mathematics & Statistics (Commerce) Part 1

Find the area of region bounded by \(y^2 = 25x\) and the line \(x = 4\).

Q.3 (A)(iii) Find the area of region bounded by \(y^2 = 25x\) and the line \(x = 4\).

Doubtly Asked 29 Oct 2026 Board exam · July 2024 hsc 12th Save WhatsApp Report

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Answer: \(\dfrac{160}{3}\) sq. units

\(y^2 = 25x\) is a parabola with vertex at the origin, symmetric about the X-axis. The line \(x = 4\) meets it at \(y = \pm 10\), i.e. at \((4, 10)\) and \((4, -10)\).

By symmetry, required area \(= 2 \times\) (area above the X-axis), where \(y = 5\sqrt{x}\):

\(\displaystyle A = 2\int_0^4 5\sqrt{x}\,dx = 10\left[\frac{x^{3/2}}{3/2}\right]_0^4 = 10 \cdot \frac{2}{3}\left(4^{3/2} - 0\right) = \frac{20}{3} \times 8 = \frac{160}{3}\) sq. units.

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

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Find the area of region bounded by \(y^2 = 25x\) and the line \(x = 4\).

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