Question
Let $A_{1}$ be the bounded area enclosed by the curves $y = x^{2} + 2,x + y = 8$ and y-axis that lies in the first quadrant. Let $A_{2}$ be the bounded area enclosed by the curves $y = x^{2} + 2,y^{2} = x,x = 2$, and y -axis that lies in the first quadrant. Then $A_{1} - A_{2}$ is equal to
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Solution
$${A_{1} = \int_{0}^{2}\mspace{2mu}\left( (8 - x) - \left( x^{2} + 2 \right) \right)dx }{= A_{1} = \int_{0}^{2}\mspace{2mu}\left( 6 - x - x^{2} \right)dx }{A_{1}\left( 6x - \frac{x^{2}}{2} - \frac{x^{3}}{3} \right)_{0}^{2} = 12 - 2 - \frac{8}{3} = 10 - \frac{8}{3} = \frac{22}{3} }$$
$${A_{2} = \int_{0}^{2}\mspace{2mu}\left( x^{2} + 2 \right)dx - \frac{2}{3}(2\sqrt{2}) }{A_{2} = \left( \frac{x^{3}}{3} + 2x \right)_{0}^{2} - \frac{4\sqrt{2}}{3} }{A_{2} = \frac{8}{3} + 4 - \frac{4\sqrt{2}}{3} = \frac{20}{3} - \frac{4\sqrt{2}}{3} }{A_{1} - A_{2} = \frac{2}{3} + \frac{4\sqrt{2}}{3}}$$
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$$\begin{matrix} & x^{2} + 3x + 1 - x - 2 = 0 \\ & x^{2} + 2x - 1 = 0 \\ & x = - 1 + \sqrt{2}, - 1 - \sqrt{2} \\ & \ (\text{~}\text{III}\text{~}) \\ & \ - x^{2} - 3x + 1 - x - 2 = 0 \\ & x^{2} + 4x + 1 = 0 \\ & x = - 2 - \sqrt{3}, - 2 + \sqrt{3}\text{~}\text{(rejected)}\text{~} \end{matrix}$$
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