y = x , y = 1 2 x , x = 9 x = 1 2 x → x − 1 2 x = 0 → 4 − 1 2 ( 4 ) = 2 − 2 = 0 , x = 4 A = ∫ 0 4 [ x − 1 2 x ] d x + ∫ 4 9 [ 1 2 x − x ] d x = [ 2 3 x 3 2 − 1 4 x 2 ] 0 4 + [ 1 4 x 2 − 2 3 x 3 2 ] 4 9 = [ 2 3 ( 4 ) 3 2 − 1 4 ( 4 ) 2 ] − [ 0 ] + [ 1 4 ( 9 ) 2 − 2 3 ( 9 ) 3 2 ] − [ 1 4 ( 4 ) 2 − 2 3 ( 4 ) 3 2 ] = [ 16 3 − 4 ] − [ 0 ] + [ 81 4 − 18 ] − [ 4 − 16 3 ] = 4 3 + 9 4 + 4 3 = 8 3 + 9 4 = 32 12 + 27 12 = 59 12 {\displaystyle {\begin{aligned}y={\sqrt {x}},\ y={\frac {1}{2}}x,\ x=9\\{\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\\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\\=\ \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}\\=\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]\\={\frac {4}{3}}+{\frac {9}{4}}+{\frac {4}{3}}={\frac {8}{3}}+{\frac {9}{4}}={\frac {32}{12}}+{\frac {27}{12}}\\={\frac {59}{12}}\end{aligned}}}
![{\displaystyle {\begin{aligned}y={\sqrt {x}},\ y={\frac {1}{2}}x,\ x=9\\{\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\\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\\=\ \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}\\=\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]\\={\frac {4}{3}}+{\frac {9}{4}}+{\frac {4}{3}}={\frac {8}{3}}+{\frac {9}{4}}={\frac {32}{12}}+{\frac {27}{12}}\\={\frac {59}{12}}\end{aligned}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/51f1946caeb8dae0ca7527b0a401efedaf938bcb)