y = 1 + x y = 3 + x 3 {\displaystyle {\begin{aligned}&\color {red}\mathbf {y=1+{\sqrt {x}}} &\color {royalblue}\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 = ∫ − 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 _{0}^{9}\left(1+{\sqrt {x}}-{\frac {3+x}{3}}\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}}}
