Q.2 (B) (ii) Evaluate: ∫₂⁵√(x)/√(x)+√(7-x) dx

Answer: 3/2 Let I=∫₂⁵√(x)/√(x)+√(7-x) dx ... (1) Using ∫_a^b f(x) dx=∫_a^b f(a+b-x) dx with a+b=7 : I=∫₂⁵√(7-x)/√(7-x)+√(x) dx ... (2) Adding (1) and (2): 2I=∫₂⁵√(x)+√(7-x)/√(x)+√(7-x) dx=∫₂⁵ 1 dx=[x]₂⁵=3 . therefore I=3/2 .

Evaluate: \(\int_{2}^{5}\frac{\sqrt{x}}{\sqrt{x}+\sqrt{7-x}}\,dx\)

Q.2 (B) (ii) Evaluate: \(\int_{2}^{5}\frac{\sqrt{x}}{\sqrt{x}+\sqrt{7-x}}\,dx\)

Doubtly Asked 28 Oct 2026 Board exam · June 2026 hsc 12th Save WhatsApp Report

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Answer: \(\frac{3}{2}\)

Let \(I=\int_{2}^{5}\frac{\sqrt{x}}{\sqrt{x}+\sqrt{7-x}}\,dx\) ... (1)

Using \(\int_a^b f(x)\,dx=\int_a^b f(a+b-x)\,dx\) with \(a+b=7\):

\(I=\int_{2}^{5}\frac{\sqrt{7-x}}{\sqrt{7-x}+\sqrt{x}}\,dx\) ... (2)

Adding (1) and (2): \(2I=\int_{2}^{5}\frac{\sqrt{x}+\sqrt{7-x}}{\sqrt{x}+\sqrt{7-x}}\,dx=\int_2^5 1\,dx=[x]_2^5=3\).

\(\therefore I=\frac{3}{2}\).

Doubtly AI · 28 Oct 2026 AI-verified · Oct 2026 Report

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Evaluate: \(\int_{2}^{5}\frac{\sqrt{x}}{\sqrt{x}+\sqrt{7-x}}\,dx\)

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