6.1 Areas Between Curves/12

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

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

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

Failed to parse (syntax error): {\displaystyle \int_{0}^{2} \left[(4x-x^2) - (x^2)\right]dx = \int_{0}^{2}(4x)dx &= \left[\frac{4x^2}{2}\right]\Bigg|_{0}^{2} \\[2ex] &= \left[2x^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{align} }

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