y = x 2 y = 4 x − x 2 x = 0 x = 2 {\displaystyle {\begin{aligned}&\color {red}\mathbf {y=x^{2}} &\color {royalblue}\mathbf {y=4x-x^{2}} \\&x=0&x=2\\\end{aligned}}}
∫ 0 2 [ ( 4 x − x 2 ) − ( x 2 ) ] d x {\displaystyle \int _{0}^{2}\left[(4x-x^{2})-(x^{2})\right]dx}
4 − x 2 = x 2 4 x − 2 x 2 = 0 2 x ( 2 − x ) = 0 x = 0 x = 2 {\displaystyle {\begin{aligned}4-x^{2}&=x^{2}\\4x-2x^{2}&=0\\2x(2-x)&=0\\x&=0&x=2\end{aligned}}}
∫ 0 2 [ ( 4 x − x 2 ) − ( x 2 ) ] d x = ∫ 0 2 ( 4 x ) d x {\displaystyle \int _{0}^{2}\left[(4x-x^{2})-(x^{2})\right]dx=\int _{0}^{2}(4x)dx}
∫ − 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 = 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}}\\[2ex]&={\frac {64}{3}}\end{aligned}}}
![{\displaystyle \int _{0}^{2}\left[(4x-x^{2})-(x^{2})\right]dx}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/6d1ebe4106c00946ea1b68c9603574e2f1fdbcec)
![{\displaystyle \int _{0}^{2}\left[(4x-x^{2})-(x^{2})\right]dx=\int _{0}^{2}(4x)dx}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/0406095dc5c7b6b0a0ae7a38f759508615fe8bb2)
![{\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}}\\[2ex]&={\frac {64}{3}}\end{aligned}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/b34bbaa4ca1bd3009f9c8fd2549e9b85ac66f093)