The area (in square units) of the region boun | vedclass

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(C) The given equations are:$x^2 = 8y$  $(i)$$x = 4$  (ii)$y = 0$ (the $X$-axis)  (iii)To find the area,we integrate $y$ with respect to $x$ from $x=0

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$\frac{8}{3}$

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(C) The given equations are:$x^2 = 8y$  $(i)$$x = 4$  (ii)$y = 0$ (the $X$-axis)  (iii)To find the area,we integrate $y$ with respect to $x$ from $x=0$ to $x=4$:Area $= \int_{0}^{4} y \, dx$From $(i)$,$y = \frac{x^2}{8}$.Area $= \int_{0}^{4} \frac{x^2}{8} \, dx$$= \frac{1}{8} \left[ \frac{x^3}{3} ight]_{0}^{4}$$= \frac{1}{8} \left( \frac{4^3}{3} - \frac{0^3}{3} ight)$$= \frac{1}{8} \left( \frac{64}{3} ight)$$= \frac{8}{3} 	ext{ square units.}$

The area (in square units) of the region bounded by $x^2=8y$,$x=4$ and the $X$-axis is:

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