The area of the region bounded by $y = x^{2} + 2$ , $y = - x$ , $x = 0$ and $x = 1$ is

\frac{17}{6}

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\frac{17}{3}

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\frac{10}{3}

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\frac{8}{3}

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Solution

The correct option is A $\frac{17}{6}$
$A = \int_{0}^{1} x + 2 + x^{2} d x A = {[\frac{x^{2}}{2} + 2 x + \frac{x^{3}}{3}]}_{0}^{1} A = \frac{1}{2} + 2 + \frac{1}{3} A = \frac{17}{6}$

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