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}}}
∫ − 3 3 | ( 8 − x 2 ) − ( x 2 ) | d x {\displaystyle \int _{-3}^{3}\left|(8-x^{2})-(x^{2})\right|dx}
8 − x 2 = x 2 − 2 x 2 = − 8 x 2 = 4 x = ± 2 {\displaystyle {\begin{aligned}8-x^{2}&=x^{2}\\-2x^{2}&=-8\\x^{2}&=4\\x&=\pm 2\end{aligned}}}
∫ − 3 3 | ( 8 − x 2 ) − ( x 2 ) | d x = ∫ − 3 − 2 ( ( x 2 ) − ( 8 − x 2 ) ) d x + ∫ − 2 2 ( ( 8 − x 2 ) − ( x 2 ) ) d x + ∫ 2 3 ( ( x 2 ) − ( 8 − x 2 ) ) d x = 14 3 + 64 3 + 14 3 = 92 3 {\displaystyle \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}}}
∫ − 3 − 2 ( ( x 2 ) − ( 8 − x 2 ) ) d x = ∫ − 3 − 2 ( 2 x 2 − 8 ) ) d x = [ 2 x 3 3 − 8 x ] | − 3 − 2 = [ 2 ( − 2 ) 3 3 − 8 ( − 2 ) ] − [ 2 ( − 3 ) 3 3 − 8 ( − 3 ) ] = [ − 16 3 + 16 ] − [ − 54 3 + 24 ] = 38 3 − 8 = 14 3 {\displaystyle {\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}}}
∫ − 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 _{-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}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/eadc7ac3e6b12462f01790319a93465c651be45b)
![{\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)
