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Q.
The area (in sq units) of the region bounded by the curve $y=x^2$ and the line $y=16$ is
Application of Integrals
Solution:
Area of region $=2 \int\limits_{10}^{16} \sqrt{y} d y$
(putting $y=16$ in $y=x^2$, we get $x=\pm 4$ )
$=2\left[\frac{y^{3 / 2}}{3 / 2}\right]_0^{16}=\frac{2}{3} \times 2\left[16^{3 / 2}-0\right]$
$=\frac{4 \times 64}{3}=\frac{256}{3}$ sq units
