6.1 Areas Between Curves/18

(not done)

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

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

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

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

${\begin{aligned}\int _{-2}^{2}\left((8-x^{2})-(x^{2})\right)dx&=\int _{-2}^{2}\left(8-2x^{2}\right)dx\\[2ex]&=\left(8x-{\frac {2x^{3}}{3}}\right){\bigg |}_{-2}^{2}\\[2ex]&=\left(8(2)-{\frac {2(2)^{3}}{3}}\right)-\left(8(-2)-{\frac {2(-2)^{3}}{3}}\right)\\[2ex]&=\left(16-{\frac {16}{3}}\right)-\left(-16+{\frac {16}{3}}\right)=32-{\frac {32}{3}}={\frac {96}{3}}-{\frac {32}{3}}\\[2ex]&={\frac {64}{3}}\end{aligned}}$