Discussion
The value of \(\int_{0}^{3} x^{2}[x]\,dx\) is equal to
Answer: B
The value is 15.
The greatest integer \([x]\) is 0, 1 and 2 on [0, 1), [1, 2) and [2, 3).
\[ \begin{aligned} \int_{0}^{3}x^{2}[x]\,dx &= 0 + \int_{1}^{2}x^{2}dx + 2\int_{2}^{3}x^{2}dx \\ &= \tfrac{7}{3} + 2 \times \tfrac{19}{3} = \tfrac{45}{3} = 15 \end{aligned} \]
Without the factor [x] the integral would be 9. Forgetting the 2 on the last piece gives \(\frac{26}{3}\), and treating [x] as x gives \(\frac{81}{4}\).
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