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

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(B) Given the curve is ${x^2} = 4y$,which implies $y = \frac{x^2}{4}$.The region is bounded by the curve $y = \frac{x^2}{4}$,the $x$-axis $(y = 0)$,an

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$\frac{2}{3}$

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(B) Given the curve is ${x^2} = 4y$,which implies $y = \frac{x^2}{4}$.The region is bounded by the curve $y = \frac{x^2}{4}$,the $x$-axis $(y = 0)$,and the vertical line $x = 2$. Since the curve passes through the origin $(0,0)$,the limits of integration for $x$ are from $0$ to $2$.The area $A$ is given by the definite integral:$A = \int_{0}^{2} y \, dx = \int_{0}^{2} \frac{x^2}{4} \, dx$$A = \frac{1}{4} \int_{0}^{2} x^2 \, dx$$A = \frac{1}{4} \left[ \frac{x^3}{3} ight]_{0}^{2}$$A = \frac{1}{4} \left( \frac{2^3}{3} - \frac{0^3}{3} ight)$$A = \frac{1}{4} \left( \frac{8}{3} ight) = \frac{2}{3} 	ext{ square units.}$

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

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