First slide

Evaluation of definite integrals

Question

The value of $\int_{0}^{2} x^{[x^{2} + 1]} d x,$ where $[x]$ is
the greatest integer less than or equal to $x$ is

Moderate

A

2
B

8/3
C

4
D

none of these

Solution

For $x \in [0, 2], x^{2} + 1 \in [1, 5]$ we must break
$⌊ 0, 2 ⌋ = ⌊ 0, 1 ⌋ \cup [1, \sqrt{2}] \cup [\sqrt{2}, \sqrt{3}] \cup [\sqrt{3}, 2] .$
Hence $\int_{0}^{2} x^{[x^{2} + 1]} d x$
$\begin{array}{l} = \int_{0}^{1} x^{[x^{2} + 1]} d x + \int_{1}^{\sqrt{2}} x^{[x^{2} + 1]} d x + \int_{\sqrt{2}}^{\sqrt{3}} x^{[x^{2} + 1]} d x + \\ \int_{\sqrt{3}}^{2} x^{[x^{2} + 1]} d x \\ = \int_{0}^{1} x d x + \int_{1}^{\sqrt{2}} x^{2} d x + \int_{\sqrt{2}}^{\sqrt{3}} x^{3} d x + \int_{\sqrt{3}}^{2} x^{4} d x \\ = \frac{1}{2} + \frac{1}{3} [2^{3 / 2} - 1] + \frac{1}{4} [9 - 4] + \frac{1}{5} [32 - 3^{5 / 2}] \\ = \frac{469}{60} + \frac{1}{3} 2^{3 / 2} - \frac{1}{5} 3^{5 / 2 .} \end{array}$

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