y = 1 x y = x y = 1 4 x x > 0 {\displaystyle {\begin{aligned}&\color {red}\mathbf {y={\frac {1}{x}}} &\color {green}\mathbf {y=x} \\&\color {blue}\mathbf {y={\frac {1}{4}}x} \\&x>0\\\end{aligned}}}
∫ 0 1 ( x − 1 4 x ) d x + ∫ 1 2 ( 1 x − 1 4 x ) d x = ∫ 0 1 ( 3 4 x ) d x + ∫ 1 2 ( 1 x ) d x − ∫ 1 2 ( 1 4 x ) d x {\displaystyle \int _{0}^{1}\left(x-{\frac {1}{4}}x\right)dx+\int _{1}^{2}\left({\frac {1}{x}}-{\frac {1}{4}}x\right)dx=\int _{0}^{1}\left({\frac {3}{4}}x\right)dx+\int _{1}^{2}\left({\frac {1}{x}}\right)dx-\int _{1}^{2}\left({\frac {1}{4}}x\right)dx}
= [ 3 x 2 8 ] | 0 1 + [ ln | x | ] | 1 2 − [ 1 8 x 2 ] | 1 2 = [ 3 ( 1 ) 2 8 ] + [ l n | 2 | − l n | 1 | ] − [ 1 8 ( 2 ) 2 − 1 8 ( 1 ) 2 ] = 3 8 + l n | 2 | − 3 8 = l n ( 2 ) {\displaystyle {\begin{aligned}&=\left[{\frac {3x^{2}}{8}}\right]{\Bigg |}_{0}^{1}+\left[\ln |x|\right]{\Bigg |}_{1}^{2}-\left[{\frac {1}{8}}x^{2}\right]{\Bigg |}_{1}^{2}\\[2ex]&=\left[{\frac {3(1)^{2}}{8}}\right]+\left[ln|2|-ln|1|\right]-\left[{\frac {1}{8}}(2)^{2}-{\frac {1}{8}}(1)^{2}\right]\\[2ex]&={\frac {3}{8}}+ln|2|-{\frac {3}{8}}&=ln(2)\end{aligned}}}
![{\displaystyle {\begin{aligned}&=\left[{\frac {3x^{2}}{8}}\right]{\Bigg |}_{0}^{1}+\left[\ln |x|\right]{\Bigg |}_{1}^{2}-\left[{\frac {1}{8}}x^{2}\right]{\Bigg |}_{1}^{2}\\[2ex]&=\left[{\frac {3(1)^{2}}{8}}\right]+\left[ln|2|-ln|1|\right]-\left[{\frac {1}{8}}(2)^{2}-{\frac {1}{8}}(1)^{2}\right]\\[2ex]&={\frac {3}{8}}+ln|2|-{\frac {3}{8}}&=ln(2)\end{aligned}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/74245465395e36719a3629579583cd8d02883a24)
