Q.2 (B) (iii) Evaluate : ∫₂⁷√(x)/√(x)+√(9-x) dx

Answer: 5/2 Let I=∫₂⁷√(x)/√(x)+√(9-x) dx ...(1) Using ∫_a^b f(x) dx=∫_a^b f(a+b-x) dx with a+b=9 : I=∫₂⁷√(9-x)/√(9-x)+√(x) dx ...(2) Adding (1) and (2): 2I=∫₂⁷√(x)+√(9-x)/√(x)+√(9-x) dx=∫₂⁷ 1 dx=[x]₂⁷=5 . therefore I=5/2 .

Evaluate : \(\int_2^7\frac{\sqrt{x}}{\sqrt{x}+\sqrt{9-x}}\,dx\)

Q.2 (B) (iii) Evaluate : \(\int_2^7\frac{\sqrt{x}}{\sqrt{x}+\sqrt{9-x}}\,dx\)

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

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Answer: \(\frac{5}{2}\)

Let \(I=\int_2^7\frac{\sqrt{x}}{\sqrt{x}+\sqrt{9-x}}\,dx\) ...(1)

Using \(\int_a^b f(x)\,dx=\int_a^b f(a+b-x)\,dx\) with \(a+b=9\):

\(I=\int_2^7\frac{\sqrt{9-x}}{\sqrt{9-x}+\sqrt{x}}\,dx\) ...(2)

Adding (1) and (2): \(2I=\int_2^7\frac{\sqrt{x}+\sqrt{9-x}}{\sqrt{x}+\sqrt{9-x}}\,dx=\int_2^7 1\,dx=[x]_2^7=5\).

\(\therefore I=\frac{5}{2}\).

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

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Evaluate : \(\int_2^7\frac{\sqrt{x}}{\sqrt{x}+\sqrt{9-x}}\,dx\)

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