6.1 Areas Between Curves/18

(not done)

${\begin{aligned}&\color {red}\mathbf {y=8-x^{2}} &\color {royalblue}\mathbf {y=x^{2}} \\&x=-3&x=3\\\end{aligned}}$

$\int _{-3}^{3}\left|(8-x^{2})-(x^{2})\right|dx$

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

$\int _{-3}^{3}\left|(8-x^{2})-(x^{2})\right|dx=\int _{-3}^{-2}\left((x^{2})-(8-x^{2})\right)dx+\int _{-2}^{2}\left((8-x^{2})-(x^{2})\right)dx+\int _{2}^{3}\left((x^{2})-(8-x^{2})\right)dx$

Failed to parse (unknown function "\math"): {\displaystyle \begin{align} \int_{-3}^{-2}\left((x^2)-(8-x^2)\right)dx \end{align} <\math> <math> \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} }