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Evaluation of definite integrals

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Question

$\int_{0}^{2} [x^{2}] d x =$

Moderate

Solution

$\begin{array}{l} G.I = \int_{0}^{1} [x^{2}] d x + \int_{1}^{\sqrt{2}} [x^{2}] d x + \int_{\sqrt{2}}^{\sqrt{3}} [x^{2}] d x + \int_{\sqrt{3}}^{2} [x^{2}] d x \\ = \int_{0}^{1} o d x + \int_{1}^{\sqrt{2}} (1) d x + \int_{\sqrt{2}}^{\sqrt{3}} (2) d x + \int_{\sqrt{3}}^{2} (3) d x \\ = 0 + (x)_{1}^{\sqrt{2}} + 2 (x)_{\sqrt{2}}^{\sqrt{3}} + 3 (x)_{\sqrt{3}}^{2} \\ = \sqrt{2} - 1 + 2 \sqrt{3} - 2 \sqrt{2} + 6 - 3 \sqrt{3} \\ = 5 - \sqrt{2} - \sqrt{3} = 5 - 1.4142 - 1.7320 \\ = 5 - 3.1462 \\ = 1.8538 \end{array}$

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