 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2019/paper-1/6/solution

u = ln x

and

d v = x^{2} d x

. This gives the antiderivative

\int x^{2} ln x d x = \frac{x ^{3}}{3} ln x - \frac{1}{3} \int x^{2} d x = \frac{x ^{3}}{3} ln x - \frac{x ^{3}}{9} + C .

Evaluating at the two bounds, with

ln 1 = 0

, yields

\int_{1}^{y} x^{2} ln x d x = \frac{y ^{3}}{3} ln y - \frac{y ^{3} - 1}{9} .

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