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Question

Find the area of region bounded by the line $$x = 2$$ and the parabola $$y^{2} = 8x$$.


Solution

$$\begin{array}{l}\dfrac{1}{{1 + {x^2}}} - \log \left( {1 + {x^2}} \right)\\{y^2} = 8x\\y = \sqrt {8x} \\y = 2\sqrt {2x} \\{\rm{area}} = 2\int\limits_0^2 {ydx}  = 2\int\limits_0^2 {2\sqrt 2 \sqrt x dx} \\ = 4\sqrt 2 \int\limits_0^2 {\sqrt x dx}  = 4\sqrt 2 \left[ {\dfrac{{{x^{3/2}}}}{{3/2}}} \right]_0^2\\ = \dfrac{{8\sqrt 2 }}{3}\left[ {{2^{3/2}} - 0} \right] = \dfrac{{8\sqrt 2 }}{3}2\sqrt 2  = \dfrac{{32}}{3}\end{array}$$

Mathematics
RS Agarwal
Standard XII

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