y = 8 − x 2 y = x 2 x = − 3 x = 3 {\displaystyle {\begin{aligned}&\color {red}\mathbf {y=8-x^{2}} &\color {royalblue}\mathbf {y=x^{2}} \\&x=-3&x=3\\\end{aligned}}}
8 − x 2 = x 2 8 = 2 x 2 4 = x 2 {\displaystyle {\begin{aligned}8-x^{2}&=x^{2}\\8&=2x^{2}\\4&=x^{2}\\\end{aligned}}}
x = ± 2 {\displaystyle x=\pm 2}
∫ − 2 2 ( ( 8 − x 2 ) − ( x 2 ) ) d x = ∫ − 2 2 ( 8 − 2 x 2 ) d x = ( 8 x − 2 x 3 3 ) | − 2 2 = ( 8 ( 2 ) − 2 ( 2 ) 3 3 ) − ( 8 ( − 2 ) − 2 ( − 2 ) 3 3 ) = ( 16 − 16 3 ) − ( − 16 + 16 3 ) = 32 − 32 3 = 96 3 − 32 3 = 64 3 {\displaystyle {\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}}}
![{\displaystyle {\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}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/50459a9dbb29e4b32f4304fd65349f53a7bfea9e)
