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What is the area bounded by the curve $y=x^{2}$ and the line $y=16 ?$
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The correct answer is:
$256 / 3$
Given, equations of curves are $y=x^{2} \quad \ldots$ (i)
and $y=16$ $\ldots$ (ii) On solving Eqs. (i) and (ii), we get $x^{2}=16 \Rightarrow x=4,-4$
$\therefore \quad$ Points of intersection are $(4,16)$, and $(-4,16)$.
Required area $=\int_{-4}^{4}\left(16-x^{2}\right) d x=2 \int_{0}^{4}\left(16-x^{2}\right) d x$
$=2\left[16 x-\frac{x^{3}}{3}\right]_{0}^{4}=2\left[64-\frac{64}{3}\right]=2 \times 64 \times \frac{2}{3}$
$=\frac{256}{3}$ sq unit
and $y=16$ $\ldots$ (ii) On solving Eqs. (i) and (ii), we get $x^{2}=16 \Rightarrow x=4,-4$
$\therefore \quad$ Points of intersection are $(4,16)$, and $(-4,16)$.
Required area $=\int_{-4}^{4}\left(16-x^{2}\right) d x=2 \int_{0}^{4}\left(16-x^{2}\right) d x$
$=2\left[16 x-\frac{x^{3}}{3}\right]_{0}^{4}=2\left[64-\frac{64}{3}\right]=2 \times 64 \times \frac{2}{3}$
$=\frac{256}{3}$ sq unit
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