6.1 Areas Between Curves/10

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

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 \begin{align} &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{align} }

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_{-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}}