6.1 Areas Between Curves/27

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

Failed to parse (syntax error): {\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 = }

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