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

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(B) Given equations of the curve and the line are $y^{2}=8x$ and $y=2x$.To find the intersection points,substitute $y=2x$ into $y^{2}=8x$:$(2x)^{2}=8x

Vedclass · Accepted Answer

$\frac{4}{3}$ sq. units

Vedclass · Answer

(B) Given equations of the curve and the line are $y^{2}=8x$ and $y=2x$.To find the intersection points,substitute $y=2x$ into $y^{2}=8x$:$(2x)^{2}=8x$$4x^{2}=8x$$4x^{2}-8x=0$$4x(x-2)=0$So,$x=0$ and $x=2$.When $x=0$,$y=0$. When $x=2$,$y=4$.The intersection points are $(0,0)$ and $(2,4)$.The required area $A$ is given by the integral of the upper curve minus the lower curve from $x=0$ to $x=2$:$A = \int_{0}^{2} (\sqrt{8x} - 2x) dx$$A = \int_{0}^{2} (2\sqrt{2}x^{1/2} - 2x) dx$$A = 2\sqrt{2} \int_{0}^{2} x^{1/2} dx - 2 \int_{0}^{2} x dx$$A = 2\sqrt{2} \left[ \frac{x^{3/2}}{3/2} \right]_{0}^{2} - 2 \left[ \frac{x^{2}}{2} \right]_{0}^{2}$$A = 2\sqrt{2} \cdot \frac{2}{3} [x^{3/2}]_{0}^{2} - [x^{2}]_{0}^{2}$$A = \frac{4\sqrt{2}}{3} (2^{3/2}) - (2^{2})$$A = \frac{4\sqrt{2}}{3} (2\sqrt{2}) - 4$$A = \frac{4 \cdot 2 \cdot 2}{3} - 4 = \frac{16}{3} - 4 = \frac{16-12}{3} = \frac{4}{3}$ sq. units.

The area of the region bounded by the curve $y^{2}=8x$ and the line $y=2x$ is

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