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The area of the region bounded by the curve y = x 2 and the line y = 16 |
101/3 256/3 322/3 235/3 |
256/3 |
\(⇒2\int_{0}^{16}\sqrt{y}dy\) $=2. \frac{2}{3}\begin{bmatrix}y^{\frac{3}{2}}\end{bmatrix}_0^{16}$ $⇒2(\frac{2}{3})[4^3]$ $=\frac{256}{3}$ sq. units
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