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 (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle \int_{0}^{2} \left[(4x-x^2) - (x^2)\right] dx}

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle \begin{align} 4-x^2 &= x^2 \\ 4x-2x^2 &= 0 \\ 2x(2-x) &= 0 \\ x &= 0& x = 2 \end{align} }

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

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