The area bounded by the parabola y^2 = 4ax,it | vedclass

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(D) The equation of the parabola is $y^2 = 4ax$,which implies $y = \pm 2\sqrt{a}\sqrt{x}$.Since the area is bounded by the axis (x-axis) and the curve

Vedclass · Accepted Answer

$\frac{152\sqrt{a}}{3}$

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(D) The equation of the parabola is $y^2 = 4ax$,which implies $y = \pm 2\sqrt{a}\sqrt{x}$.Since the area is bounded by the axis (x-axis) and the curve,we consider the upper part of the parabola $y = 2\sqrt{a}\sqrt{x}$.The area $A$ bounded by the curve $y = f(x)$,the x-axis,and the ordinates $x = 4$ and $x = 9$ is given by the integral:$A = \int_{4}^{9} y \, dx = \int_{4}^{9} 2\sqrt{a}\sqrt{x} \, dx$$A = 2\sqrt{a} \int_{4}^{9} x^{1/2} \, dx$$A = 2\sqrt{a} \left[ \frac{x^{3/2}}{3/2} ight]_{4}^{9}$$A = 2\sqrt{a} \cdot \frac{2}{3} \left[ x^{3/2} ight]_{4}^{9}$$A = \frac{4\sqrt{a}}{3} (9^{3/2} - 4^{3/2})$$A = \frac{4\sqrt{a}}{3} (27 - 8)$$A = \frac{4\sqrt{a}}{3} (19) = \frac{76\sqrt{a}}{3}$.Note: If the question implies the total area bounded by the curve and the ordinates (both above and below the axis),the area would be $2 	imes \frac{76\sqrt{a}}{3} = \frac{152\sqrt{a}}{3}$. Given the options,the intended answer is $D$.

The area bounded by the parabola $y^2 = 4ax$,its axis and two ordinates $x = 4$ and $x = 9$ is

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