6.1 Areas Between Curves/10

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

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

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle \int_{0}^{9} \left(1+\sqrt{x} - \frac{3+x}{3}\right)dx = \int_{0}^{9} \left(1+\sqrt{x}\right - \frac{1}3\int_{0}^{9}\left(3+x\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}}2(3) - 9 = \frac{54}3 - \frac{81}6 = \frac{108}6 - \frac{81}6 = \frac{27}6 = \frac{9}2 }