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 }

${\begin{aligned}8-x^{2}&=x^{2}\\-2x^{2}&=-8\\x^{2}&=4\\x&=\pm 2\end{aligned}}$

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 (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}^{-2}\left((x^2)-(8-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_{-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{align} }