Find the area of the closed figure bounded by the following curves:
$y = 0, y = 2 x^{2} + 1, x = - 1, x = 1.$

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Solution

$y = 0, y = 2 x^{2} + 1, x = - 1, x = 1$

$\int_{- 1}^{1} y d x = \int_{- 1}^{1} 2 x^{2} + 1 d x$

$= 2 \int_{0}^{1} 2 x^{2} + 1 d x$

$= 2 [\frac{2 x^{3}}{3} + x |_{0}^{1}]$

$= 2 [\frac{2}{3} + 1]$

$= \frac{10}{3}$ is the required area.

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