6.1 Areas Between Curves/27

${\begin{aligned}&\color {red}\mathbf {y={\frac {1}{x}}} &\color {green}\mathbf {y=x} \\&\color {blue}\mathbf {y={\frac {1}{4}}x} \\&x>0\\\end{aligned}}$

$\int _{0}^{1}\left(x-{\frac {1}{4}}x\right)\,dx+\int _{1}^{2}\left|({\frac {1}{x}}-{\frac {1}{4}}x)\right|dx=\int _{0}^{1}\left|({\frac {3}{4}}x)\right|dx+\int _{1}^{2}\left|({\frac {1}{x}})\right|dx-\int _{1}^{2}\left|({\frac {1}{4}}x)\right|dx=$

${\begin{aligned}&=\left[{\frac {3x^{2}}{8}}\right]{\Bigg |}_{0}^{1}+\left[\ln(x)\right]{\Bigg |}_{1}^{2}-\left[{\frac {1}{8}}x^{2}\right]{\Bigg |}_{1}^{2}=&=\left[{\frac {3(1)^{2}}{8}}\right]+\left[ln(2)-ln(1)\right]-\left[{\frac {1}{8}}(2)^{2}-{\frac {1}{8}}(1)^{2}\right]\\[2ex]&=frac{3}{8}+ln(2)]-{\frac {3}{8}}&=ln(2)\end{aligned}}$