y = x + 1 y = 9 − x 2 x = − 1 x = 2 {\displaystyle {\begin{aligned}&\color {red}\mathbf {y=x+1} &\color {royalblue}\mathbf {y=9-x^{2}} \\&x=-1&x=2\\\end{aligned}}}
∫ − 1 2 [ ( 9 − x 2 ) − ( x + 1 ) ] d x = [ ( 9 x − x 3 3 − x 2 2 − x ) ] | − 1 2 = [ 9 ( 2 ) − ( 2 ) 3 3 − ( 2 ) 2 2 − ( 2 ) ] − [ ( 9 ( − 1 ) − ( − 1 ) 3 3 − ( − 1 ) 2 2 − ( − 1 ) ) ] = [ 18 − 8 3 − 2 − 2 ] − [ 9 − 1 3 + 1 2 − 1 ] = 19.5 {\displaystyle {\begin{aligned}\int _{-1}^{2}[(9-x^{2})-(x+1)]dx&=[(9x-{\frac {x^{3}}{3}}-{\frac {x^{2}}{2}}-x)]{\Bigg |}_{-1}^{2}\\[2ex]&=[9(2)-{\frac {(2)^{3}}{3}}-{\frac {(2)^{2}}{2}}-(2)]-[(9(-1)-{\frac {(-1)^{3}}{3}}-{\frac {(-1)^{2}}{2}}-(-1))]\\[2ex]&=[18-{\frac {8}{3}}-2-2]-[9-{\frac {1}{3}}+{\frac {1}{2}}-1]\\[2ex]&=19.5\\[2ex]\end{aligned}}}
![{\displaystyle {\begin{aligned}\int _{-1}^{2}[(9-x^{2})-(x+1)]dx&=[(9x-{\frac {x^{3}}{3}}-{\frac {x^{2}}{2}}-x)]{\Bigg |}_{-1}^{2}\\[2ex]&=[9(2)-{\frac {(2)^{3}}{3}}-{\frac {(2)^{2}}{2}}-(2)]-[(9(-1)-{\frac {(-1)^{3}}{3}}-{\frac {(-1)^{2}}{2}}-(-1))]\\[2ex]&=[18-{\frac {8}{3}}-2-2]-[9-{\frac {1}{3}}+{\frac {1}{2}}-1]\\[2ex]&=19.5\\[2ex]\end{aligned}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/6f8c49e9f2981ba00383ad9c7b4c314f77185b05)
