Q.3 (B) (i) Evaluate: ∫ x²· e^3x dx

Answer: e^3x(x²/3-2x/9+2/27)+c Integrating by parts (taking x² as the first function): I=x²·frace^3x3-∫ 2x·frace^3x3dx=fracx²e^3x3-2/3∫ x e^3xdx Again by parts: ∫ x e^3xdx=x·frace^3x3-∫frace^3x3dx=fracxe^3x3-frace^3x9 therefore I=fracx²e^3x3-2/3[fracxe^3x3-frace^3x9]+c=e^3x(x²/3-2x/9+2/27)+c

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

Evaluate: \(\int x^2\cdot e^{3x}\,dx\)

Q.3 (B) (i) Evaluate: \(\int x^2\cdot e^{3x}\,dx\)

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

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Answer: \(e^{3x}\left(\frac{x^2}{3}-\frac{2x}{9}+\frac{2}{27}\right)+c\)

Integrating by parts (taking \(x^2\) as the first function):
\(I=x^2\cdot\frac{e^{3x}}{3}-\int 2x\cdot\frac{e^{3x}}{3}dx=\frac{x^2e^{3x}}{3}-\frac{2}{3}\int x\,e^{3x}dx\)

Again by parts: \(\int x\,e^{3x}dx=x\cdot\frac{e^{3x}}{3}-\int\frac{e^{3x}}{3}dx=\frac{xe^{3x}}{3}-\frac{e^{3x}}{9}\)

\(\therefore I=\frac{x^2e^{3x}}{3}-\frac{2}{3}\left[\frac{xe^{3x}}{3}-\frac{e^{3x}}{9}\right]+c=e^{3x}\left(\frac{x^2}{3}-\frac{2x}{9}+\frac{2}{27}\right)+c\)

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

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Evaluate: \(\int x^2\cdot e^{3x}\,dx\)

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