6.1 Areas Between Curves/12

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

$\int _{0}^{2}\left[(4x-x^{2})-(x^{2})\right]dx$

${\begin{aligned}4-x^{2}&=x^{2}\\4x-2x^{2}&=0\\2x(2-x)&=0\\x&=0&x=2\end{aligned}}$

$\int _{0}^{2}\left[(4x-x^{2})-(x^{2})\right]dx=\int _{0}^{2}(4x)dx<math>{\begin{aligned}\int _{-3}^{-2}\left((x^{2})-(8-x^{2})\right)dx&=\int _{-3}^{-2}\left(2x^{2}-8)\right)dx\\[2ex]&=\left[{\frac {2x^{3}}{3}}-8x\right]{\Bigg |}_{-3}^{-2}\\[2ex]&=\left[{\frac {2(-2)^{3}}{3}}-8(-2)\right]-\left[{\frac {2(-3)^{3}}{3}}-8(-3)\right]\\[2ex]&=\left[{\frac {-16}{3}}+16\right]-\left[{\frac {-54}{3}}+24\right]={\frac {38}{3}}-8\\[2ex]&={\frac {14}{3}}\end{aligned}}$

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