6.1 Areas Between Curves/17

${\begin{aligned}&\color {red}\mathbf {y={\sqrt {x}}} ,\ \color {blue}\mathbf {y={\frac {1}{2}}x} ,\ \color {green}\mathbf {x=9} \\[2ex]&{\sqrt {x}}={\frac {1}{2}}x\ \rightarrow \ {\sqrt {x}}-{\frac {1}{2}}x=0\ \rightarrow \ {\sqrt {4}}\ -{\frac {1}{2}}(4)=2-2=0,\ x=4\\[2ex]&A=\int _{0}^{4}\left[{\sqrt {x}}-{\frac {1}{2}}x\right]\mathrm {d} x+\int _{4}^{9}\left[{\frac {1}{2}}x-{\sqrt {x}}\right]\mathrm {d} x\\[2ex]&=\ \left[{\frac {2}{3}}x^{\frac {3}{2}}-{\frac {1}{4}}x^{2}\right]_{0}^{4}\ +\ \left[{\frac {1}{4}}x^{2}-{\frac {2}{3}}x^{\frac {3}{2}}\right]_{4}^{9}\\[2ex]&=\left[{\frac {2}{3}}\left(4\right)^{\frac {3}{2}}-{\frac {1}{4}}\left(4\right)^{2}\right]-\left[0\right]+\left[{\frac {1}{4}}\left(9\right)^{2}-{\frac {2}{3}}\left(9\right)^{\frac {3}{2}}\right]-\left[{\frac {1}{4}}\left(4\right)^{2}-{\frac {2}{3}}\left(4\right)^{\frac {3}{2}}\right]=\left[{\frac {16}{3}}-4\right]-\left[0\right]+\left[{\frac {81}{4}}-18\right]-\left[4-{\frac {16}{3}}\right]\\[2ex]&={\frac {4}{3}}+{\frac {9}{4}}+{\frac {4}{3}}={\frac {8}{3}}+{\frac {9}{4}}={\frac {32}{12}}+{\frac {27}{12}}\\[2ex]&={\frac {59}{12}}\end{aligned}}$