Q.3 (B) (ii) Evaluate: ∫₁⁴fracsqrt[3]x+6sqrt[3]x+6+sqrt[3]11-x dx

Answer: 3/2 Let I=∫₁⁴fracsqrt[3]x+6sqrt[3]x+6+sqrt[3]11-xdx ... (1) Using ∫_a^b f(x)dx=∫_a^b f(a+b-x)dx with a+b=5: x+6→ 11-x and 11-x→ x+6. I=∫₁⁴fracsqrt[3]11-xsqrt[3]11-x+sqrt[3]x+6dx ... (2) Adding (1) and (2): 2I=∫₁⁴fracsqrt[3]x+6+sqrt[3]11-xsqrt[3]x+6+sqrt[3]11-xdx=∫₁⁴ 1 dx=[x]₁⁴=3 therefore I=3/2.

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

Evaluate: \(\int_1^4\frac{\sqrt[3]{x+6}}{\sqrt[3]{x+6}+\sqrt[3]{11-x}}\,dx\)

Q.3 (B) (ii) Evaluate: \(\int_1^4\frac{\sqrt[3]{x+6}}{\sqrt[3]{x+6}+\sqrt[3]{11-x}}\,dx\)

Doubtly Asked 29 Oct 2026 Board exam · July 2025 hsc 12th Save WhatsApp Report

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

Let \(I=\int_1^4\frac{\sqrt[3]{x+6}}{\sqrt[3]{x+6}+\sqrt[3]{11-x}}dx\) ... (1)

Using \(\int_a^b f(x)dx=\int_a^b f(a+b-x)dx\) with \(a+b=5\): \(x+6\to 11-x\) and \(11-x\to x+6\).

\(I=\int_1^4\frac{\sqrt[3]{11-x}}{\sqrt[3]{11-x}+\sqrt[3]{x+6}}dx\) ... (2)

Adding (1) and (2): \(2I=\int_1^4\frac{\sqrt[3]{x+6}+\sqrt[3]{11-x}}{\sqrt[3]{x+6}+\sqrt[3]{11-x}}dx=\int_1^4 1\,dx=[x]_1^4=3\)

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

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

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Evaluate: \(\int_1^4\frac{\sqrt[3]{x+6}}{\sqrt[3]{x+6}+\sqrt[3]{11-x}}\,dx\)

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