The area of the region bounded by the curve y | vedclass

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(A) To find the area of the region bounded by the curve $y^2=4x$ and the line $y=x$,we first determine the points of intersection by substituting $y=x

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$\frac{8}{3} 	ext{ sq. units}$

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(A) To find the area of the region bounded by the curve $y^2=4x$ and the line $y=x$,we first determine the points of intersection by substituting $y=x$ into the equation of the parabola: $x^2 = 4x$ $x^2 - 4x = 0$ $x(x-4) = 0$ Thus,the points of intersection are $x=0$ and $x=4$. For $x=0$,$y=0$,so the origin $O(0,0)$ is one point. For $x=4$,$y=4$,so the point $P(4,4)$ is the other point. The required area $A$ is given by the integral of the upper curve minus the lower curve from $x=0$ to $x=4$: $A = \int_0^4 (\sqrt{4x} - x) dx$ $A = 2 \int_0^4 x^{1/2} dx - \int_0^4 x dx$ $A = 2 \left[ \frac{x^{3/2}}{3/2} ight]_0^4 - \left[ \frac{x^2}{2} ight]_0^4$ $A = 2 \cdot \frac{2}{3} [x^{3/2}]_0^4 - \left[ \frac{16}{2} - 0 ight]$ $A = \frac{4}{3} (4^{3/2}) - 8$ $A = \frac{4}{3} (8) - 8$ $A = \frac{32}{3} - 8 = \frac{32-24}{3} = \frac{8}{3} 	ext{ sq. units.}$

The area of the region bounded by the curve $y^2=4x$ and the line $y=x$ is

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