The area bounded by $y = x^{2} + 3$ and $y = 2 x + 3$ is (in sq. units)

\frac{12}{7}

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\frac{4}{3}

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\frac{3}{4}

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\frac{8}{3}

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\frac{3}{8}

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Solution

The correct option is A $\frac{4}{3}$
Required area $= \int_{0}^{2} (2 x + 3) - (x^{2} + 3) d x$
$= \int_{0}^{2} (2 x - x^{2}) d x$
$= {[\frac{2 x^{2}}{2} - \frac{x^{3}}{3}]}_{0}^{2}$
$= [4 - \frac{8}{3}] = \frac{4}{3}$ sq units

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