6.1 Areas Between Curves/12

${\begin{aligned}&\color {red}\mathbf {y=x^{2}} &\color {royalblue}\mathbf {y=4x-x^{2}} \\&x=0&x=2\\\end{aligned}}$

Failed to parse (syntax error): {\displaystyle \int_{0}^{2} \left[(4x-x^2) - (x^2)^2\]right dx }

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

$\int _{-3}^{3}\left|(8-x^{2})-(x^{2})\right|dx=\int _{-3}^{-2}\left((x^{2})-(8-x^{2})\right)dx+\int _{-2}^{2}\left((8-x^{2})-(x^{2})\right)dx+\int _{2}^{3}\left((x^{2})-(8-x^{2})\right)dx={\frac {14}{3}}+{\frac {64}{3}}+{\frac {14}{3}}={\frac {92}{3}}$

${\begin{aligned}\int _{-3}^{-2}\left((x^{2})-(8-x^{2})\right)dx&=\int _{-3}^{-2}\left(2x^{2}-8)\right)dx\\[2ex]&=\left[{\frac {2x^{3}}{3}}-8x\right]{\Bigg |}_{-3}^{-2}\\[2ex]&=\left[{\frac {2(-2)^{3}}{3}}-8(-2)\right]-\left[{\frac {2(-3)^{3}}{3}}-8(-3)\right]\\[2ex]&=\left[{\frac {-16}{3}}+16\right]-\left[{\frac {-54}{3}}+24\right]={\frac {38}{3}}-8\\[2ex]&={\frac {14}{3}}\end{aligned}}$

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