6.1 Areas Between Curves/12

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle \begin{align} & \color{red}\mathbf{y=x^2} & \color{royalblue}\mathbf{y=4x-x^2} \\ & x=0 & x=2 \\ \end{align} }

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle \int_{0}^{2} \left[(4x-x^2) - (x^2)^2\right] dx }

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle \begin{align} 4-x^2 &= x^2 \\ 4x-2x^2 &= 0 \\ 2x(2-x) &= 0 \\ x &= 0& x = 2 \end{align} }

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle \int_{0}^{2} \left[(4x-x^2) - (x^2)^2\right]dx = \int_{0}^{2} \left[(16x^2+x^4-8x^3) - (x^4)\right]dx = \int_{0}^{2} \left[(16x^2-8x^3)\right]dx = \frac{92}{3}}

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle \begin{align} \int_{-3}^{-2}\left((x^2)-(8-x^2)\right)dx &= \int_{-3}^{-2}\left(2x^2-8)\right)dx \\[2ex] &= \left[\frac{2x^3}{3}-8x\right]\Bigg|_{-3}^{-2} \\[2ex] &= \left[\frac{2(-2)^3}{3}-8(-2)\right]-\left[\frac{2(-3)^3}{3}-8(-3)\right] \\[2ex] &= \left[\frac{-16}{3}+16\right]-\left[\frac{-54}{3}+24\right] = \frac{38}{3}-8 \\[2ex] &= \frac{14}{3} \end{align} }

${\begin{aligned}\int _{-2}^{2}\left((8-x^{2})-(x^{2})\right)dx&=\int _{-2}^{2}\left(8-2x^{2}\right)dx\\[2ex]&=\left[8x-{\frac {2x^{3}}{3}}\right]{\Bigg |}_{-2}^{2}\\[2ex]&=\left[8(2)-{\frac {2(2)^{3}}{3}}\right]-\left[8(-2)-{\frac {2(-2)^{3}}{3}}\right]\\[2ex]&=\left[16-{\frac {16}{3}}\right]-\left[-16+{\frac {16}{3}}\right]=32-{\frac {32}{3}}\\[2ex]&={\frac {64}{3}}\end{aligned}}$