First slide

Evaluation of definite integrals

Question

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

Difficult

A

$2 - \sqrt{2}$
B

$2 + \sqrt{2}$
C

$\sqrt{2} - 1$
D

$\sqrt{2} - 2$

Solution

$\int_{0}^{\sqrt{2}} [x^{2}] d x = \int_{0}^{1} [x^{2}] d x + \int_{1}^{\sqrt{2}} [x^{2}] d x$
$= \int_{0}^{1} 0 \cdot d x + \int_{1}^{\sqrt{2}} 1 \cdot d x = \sqrt{2} - 1$

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