y = 1 x y = x y = 1 4 x x > 0 {\displaystyle {\begin{aligned}&\color {red}\mathbf {y={\frac {1}{x}}} &\color {royalblue}\mathbf {y=x} \\&\color {green}\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 = ∫ 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|d=\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}=&=\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]&=\left[{\frac {3}{8}}\right]+\left[ln(2)\right]-\left[{\frac {3}{8}}\right]&=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}=&=\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]&=\left[{\frac {3}{8}}\right]+\left[ln(2)\right]-\left[{\frac {3}{8}}\right]&=ln(2)\end{aligned}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/96f88cc697c4ef2ab38115746acd892edb79c3a7)
