The area (in sq. units) of the region bounded by the parabola, $y = x^{2} + 2$ and the lines, $y = x + 1, x = 0$ and $x = 3,$ is :

\frac{15}{4}

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\frac{15}{2}

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\frac{21}{2}

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\frac{17}{4}

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Solution

The correct option is B $\frac{15}{2}$
Req. area $= \int_{0}^{3} (x^{2} + 2) d x - \frac{1}{2} \cdot 5.3 = 9 + 6 - \frac{15}{2} = \frac{15}{2}$

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