6.1 Areas Between Curves/10

${\begin{aligned}&\color {green}\mathbf {y=1+{\sqrt {x}}} &\color {purple}\mathbf {y={\frac {3+x}{3}}} \\\\\end{aligned}}$

${\begin{aligned}&1+{\sqrt {x}}={\frac {3+x}{3}}\\&1+{\sqrt {x}}-{\frac {3+x}{3}}=0\\&{\frac {3+3{\sqrt {x}}}{3}}-{\frac {3+x}{3}}=0\\&3+3{\sqrt {x}}-3+x=0\\&3{\sqrt {x}}+x=0\\&3{\sqrt {x}}=-x\\&9x=x^{2}\\&9x-x^{2}=0\\&x(9-x)=0\\&x=0,9\end{aligned}}$

$\int _{0}^{9}\left(1+{\sqrt {x}}-{\frac {3+x}{3}}\right)dx=x+{\frac {2x^{\frac {3}{2}}}{3}}{\Bigg |}_{0}^{9}-{\frac {1}{3}}\int _{0}^{9}3+xdx=x+{\frac {2x^{\frac {3}{2}}}{3}}{\Bigg |}_{0}^{9}-{\frac {1}{3}}({\frac {x^{2}}{3}}+3x){\Bigg |}_{0}^{9}=9+{\frac {2(27)}{3}}-{\frac {9^{2}}{2}}(3)-9={\frac {54}{3}}-{\frac {81}{6}}={\frac {108}{6}}-{\frac {81}{6}}={\frac {27}{6}}={\frac {9}{2}}$