The area bounded by the parabola y^2 = 2x and | vedclass

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(B) The parabola is $y^2 = 2x$,which implies $y = \pm \sqrt{2x}$.Since the area is symmetric about the $x$-axis,the total area is twice the area above

Vedclass · Accepted Answer

$\frac{28\sqrt{2}}{3} 	ext{ sq. unit}$

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(B) The parabola is $y^2 = 2x$,which implies $y = \pm \sqrt{2x}$.Since the area is symmetric about the $x$-axis,the total area is twice the area above the $x$-axis.Area $= 2 \int_{1}^{4} y \, dx = 2 \int_{1}^{4} \sqrt{2x} \, dx$Area $= 2\sqrt{2} \int_{1}^{4} x^{1/2} \, dx$Area $= 2\sqrt{2} \left[ \frac{x^{3/2}}{3/2} ight]_{1}^{4}$Area $= 2\sqrt{2} 	imes \frac{2}{3} \left[ x^{3/2} ight]_{1}^{4}$Area $= \frac{4\sqrt{2}}{3} (4^{3/2} - 1^{3/2})$Area $= \frac{4\sqrt{2}}{3} (8 - 1) = \frac{4\sqrt{2}}{3} 	imes 7 = \frac{28\sqrt{2}}{3} 	ext{ sq. unit}$.

The area bounded by the parabola $y^2 = 2x$ and the ordinates $x = 1$ and $x = 4$ is:

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