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Date November 2008 Marks available 5 Reference code 08N.1.hl.TZ0.5
Level HL only Paper 1 Time zone TZ0
Command term Calculate Question number 5 Adapted from N/A

Question

Calculate the exact value of \(\int_1^{\text{e}} {{x^2}\ln x{\text{d}}x} \) .

Markscheme

Recognition of integration by parts     M1

\(\int {{x^2}\ln x{\text{d}}x = \left[ {\frac{{{x^3}}}{3}\ln x} \right] - \int {\frac{{{x^3}}}{3} \times \frac{1}{x}{\text{d}}x} } \)     A1A1

\( = \left[ {\frac{{{x^3}}}{3}\ln x} \right] - \int {\frac{{{x^2}}}{3}{\text{d}}x} \)

\( = \left[ {\frac{{{x^3}}}{3}\ln x - \frac{{{x^3}}}{9}} \right]\)     A1

\( \Rightarrow \int_1^{\text{e}} {{x^2}\ln x{\text{d}}x}  = \left( {\frac{{{{\text{e}}^3}}}{3} - \frac{{{{\text{e}}^3}}}{9}} \right) - \left( {0 - \frac{1}{9}} \right)\,\,\,\,\,\left( { = \frac{{2{{\text{e}}^3} + 1}}{9}} \right)\)     A1

[5 marks]

Examiners report

Most candidates recognised that a method of integration by parts was appropriate for this question. However, although a good number of correct answers were seen, a number of candidates made algebraic errors in the process. A number of students were also unable to correctly substitute the limits.

Syllabus sections

Topic 6 - Core: Calculus » 6.7 » Integration by parts.

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