6.1 Areas Between Curves/18

(not done)

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=8-x^2} & \color{royalblue}\mathbf{y=x^2} \\ & x=-3 & x=3 \\ \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_{-3}^{3} \left|(8-x^2) - (x^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} \int_{-3}^{-2}\left((x^2)-(8-x^2)\right)dx \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 \begin{align} 8-x^2 &= x^2 \\ -2x^2 &= -8 \\ x^2 &= 4 \\ x &= \pm2 \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_{-3}^{3} \left|(8-x^2) - (x^2)\right|dx = \int_{-3}^{-2}\left((x^2)-(8-x^2)\right)dx + \int_{-2}^{2} \left((8-x^2) - (x^2)\right)dx + \int_{2}^{3}\left((x^2)-(8-x^2)\right)dx}

Failed to parse (Conversion error. Server ("https://en.wikipedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\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}}={\frac {96}{3}}-{\frac {32}{3}}\\[2ex]&={\frac {64}{3}}\end{aligned}}}