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 (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 \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 }

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} \\[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{align} }