6.1 Areas Between Curves/27

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

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

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle \begin{align} &= \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{align} }