Evaluation of definite integrals

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Question

$\int_{0}^{21} [x]^{4} d x$ is equal

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Solution

$\begin{array}{l} \int_{0}^{21} [x]^{4} d x = \sum_{r = 0}^{20} \int_{r}^{r + 1} [x]^{4} d x \\ = \sum_{r = 0}^{20} r^{4} = {(\frac{20 \times 21}{2})}^{2} = 44100 \end{array}$

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