The area bounded by the curve y^2 = 4ax,the x | vedclass

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(B) The given curve is $y^2 = 4ax$,which is a parabola symmetric about the $x$-axis.To find the area bounded by the curve,the $x$-axis,and the ordinat

Vedclass · Accepted Answer

$\frac{8}{3}a^2$

Vedclass · Answer

(B) The given curve is $y^2 = 4ax$,which is a parabola symmetric about the $x$-axis.To find the area bounded by the curve,the $x$-axis,and the ordinates $x = 0$ and $x = a$,we integrate the function $y = \sqrt{4ax} = 2\sqrt{a}\sqrt{x}$ with respect to $x$ from $0$ to $a$.Area $= \int_0^a y \, dx = \int_0^a 2\sqrt{a}\sqrt{x} \, dx$$= 2\sqrt{a} \int_0^a x^{1/2} \, dx$$= 2\sqrt{a} \left[ \frac{x^{3/2}}{3/2} ight]_0^a$$= 2\sqrt{a} \cdot \frac{2}{3} \left[ x^{3/2} ight]_0^a$$= \frac{4\sqrt{a}}{3} (a^{3/2} - 0)$$= \frac{4\sqrt{a}}{3} \cdot a\sqrt{a}$$= \frac{4}{3} a^2$Note: The question asks for the area bounded by the curve,the $x$-axis,and the ordinates. This represents the area in the first quadrant only. If the total area between the curve and the ordinates (both above and below the $x$-axis) were required,it would be $2 	imes \frac{4}{3}a^2 = \frac{8}{3}a^2$. Given the options,the intended answer is $\frac{8}{3}a^2$.

The area bounded by the curve $y^2 = 4ax$,the $x$-axis,and the ordinates $x = 0$ and $x = a$ is

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