y = 1 + x y = 3 + x 3 {\displaystyle {\begin{aligned}&\color {green}\mathbf {y=1+{\sqrt {x}}} &\color {purple}\mathbf {y={\frac {3+x}{3}}} \\\\\end{aligned}}}
1 + x = 3 + x 3 1 + x − 3 + x 3 = 0 3 + 3 x 3 − 3 + x 3 = 0 3 + 3 x − 3 + x = 0 3 x + x = 0 3 x = − x 9 x = x 2 9 x − x 2 = 0 x ( 9 − x ) = 0 x = 0 , 9 {\displaystyle {\begin{aligned}&1+{\sqrt {x}}={\frac {3+x}{3}}\\&1+{\sqrt {x}}-{\frac {3+x}{3}}=0\\&{\frac {3+3{\sqrt {x}}}{3}}-{\frac {3+x}{3}}=0\\&3+3{\sqrt {x}}-3+x=0\\&3{\sqrt {x}}+x=0\\&3{\sqrt {x}}=-x\\&9x=x^{2}\\&9x-x^{2}=0\\&x(9-x)=0\\&x=0,9\end{aligned}}}
∫ 0 9 ( 1 + x − 3 + x 3 ) d x = [ x + 2 x 3 2 3 ] | 0 9 − 1 3 ∫ 0 9 ( 3 + x ) d x = [ x + 2 x 3 2 3 ] | 0 9 − [ 1 3 ( x 2 3 + 3 x ) ] | 0 9 = 9 + 2 ( 27 ) 3 − 9 2 6 − 9 = 54 3 − 81 6 = 108 6 − 81 6 = 27 6 = 9 2 {\displaystyle \int _{0}^{9}\left(1+{\sqrt {x}}-{\frac {3+x}{3}}\right)dx=\left[x+{\frac {2x^{\frac {3}{2}}}{3}}\right]{\Bigg |}_{0}^{9}-{\frac {1}{3}}\int _{0}^{9}\left(3+x\right)dx=\left[x+{\frac {2x^{\frac {3}{2}}}{3}}\right]{\Bigg |}_{0}^{9}-\left[{\frac {1}{3}}({\frac {x^{2}}{3}}+3x)\right]{\Bigg |}_{0}^{9}=9+{\frac {2(27)}{3}}-{\frac {9^{2}}{6}}-9={\frac {54}{3}}-{\frac {81}{6}}={\frac {108}{6}}-{\frac {81}{6}}={\frac {27}{6}}={\frac {9}{2}}}
![{\displaystyle \int _{0}^{9}\left(1+{\sqrt {x}}-{\frac {3+x}{3}}\right)dx=\left[x+{\frac {2x^{\frac {3}{2}}}{3}}\right]{\Bigg |}_{0}^{9}-{\frac {1}{3}}\int _{0}^{9}\left(3+x\right)dx=\left[x+{\frac {2x^{\frac {3}{2}}}{3}}\right]{\Bigg |}_{0}^{9}-\left[{\frac {1}{3}}({\frac {x^{2}}{3}}+3x)\right]{\Bigg |}_{0}^{9}=9+{\frac {2(27)}{3}}-{\frac {9^{2}}{6}}-9={\frac {54}{3}}-{\frac {81}{6}}={\frac {108}{6}}-{\frac {81}{6}}={\frac {27}{6}}={\frac {9}{2}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/ee391f39bf7fea74b7cce4681d6e5b12db2d6b79)
