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(D) The curves are $y^2 = 8x$ and $y = x$. The intersection points are found by substituting $y = x$ into $y^2 = 8x$,which gives $x^2 = 8x$,so $x(x -

Vedclass · Accepted Answer

$22$

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(D) The curves are $y^2 = 8x$ and $y = x$. The intersection points are found by substituting $y = x$ into $y^2 = 8x$,which gives $x^2 = 8x$,so $x(x - 8) = 0$. Thus,the intersection points are $(0, 0)$ and $(8, 8)$.We are looking for the area bounded by $y^2 = 8x$,$y = x$,and the line $x = 2$ in the first quadrant.At $x = 2$,the curve $y^2 = 8x$ gives $y = \sqrt{16} = 4$ (since it is in the first quadrant),and the line $y = x$ gives $y = 2$.The area $\alpha$ is given by the integral of the upper curve minus the lower curve from $x = 2$ to $x = 8$:$\alpha = \int_{2}^{8} (\sqrt{8x} - x) \, dx$$\alpha = \int_{2}^{8} (2\sqrt{2} \cdot x^{1/2} - x) \, dx$$\alpha = \left[ 2\sqrt{2} \cdot \frac{x^{3/2}}{3/2} - \frac{x^2}{2} \right]_{2}^{8}$$\alpha = \left[ \frac{4\sqrt{2}}{3} x^{3/2} - \frac{x^2}{2} \right]_{2}^{8}$$\alpha = \left( \frac{4\sqrt{2}}{3} \cdot (8)^{3/2} - \frac{8^2}{2} \right) - \left( \frac{4\sqrt{2}}{3} \cdot (2)^{3/2} - \frac{2^2}{2} \right)$$\alpha = \left( \frac{4\sqrt{2}}{3} \cdot 16\sqrt{2} - 32 \right) - \left( \frac{4\sqrt{2}}{3} \cdot 2\sqrt{2} - 2 \right)$$\alpha = \left( \frac{128}{3} - 32 \right) - \left( \frac{16}{3} - 2 \right)$$\alpha = \frac{128}{3} - 32 - \frac{16}{3} + 2 = \frac{112}{3} - 30 = \frac{112 - 90}{3} = \frac{22}{3}$Therefore,$3\alpha = 3 \cdot \frac{22}{3} = 22$.

Let $\alpha$ be the area of the region bounded by the curve $y^2 = 8x$ and the lines $y = x$ and $x = 2$,which lies in the first quadrant. Then the value of $3\alpha$ is equal to $..............$.

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