6.1 Areas Between Curves/15

${\begin{aligned}&\color {red}\mathbf {y=\tan(x)} &\color {royalblue}\mathbf {y=2\sin(x)} \\&x=-{\frac {\pi }{3}}&x={\frac {\pi }{3}}\\\end{aligned}}$

$\int _{-{\frac {\pi }{3}}}^{\frac {\pi }{3}}\left[(\tan(x))-(2\sin(x))\right]dx$

${\begin{aligned}\tan(x)&=2\sin(x)\\\tan(x)-2\sin(x)&=0\\x&=0\\\end{aligned}}$

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_{-\frac{\pi}{3}}^{\frac{\pi}{3}} \left[(\tan(x)) - (2\sin(x))\right]dx = \int_{-\frac{\pi}{3}}^{0}\left[(\tan(x)) - (2\sin(x))\right]dx + \int_{0}^{\frac{\pi}{3}} \left[(2\sin(x)) - (\tan(x))\right]dx = \frac{14}{3} + \frac{64}{3} + \frac{14}{3} = \frac{92}{3}}

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} \int_{-3}^{-2}\left((x^2)-(8-x^2)\right)dx &= \int_{-3}^{-2}\left(2x^2-8)\right)dx \\[2ex] &= \left[\frac{2x^3}{3}-8x\right]\Bigg|_{-3}^{-2} \\[2ex] &= \left[\frac{2(-2)^3}{3}-8(-2)\right]-\left[\frac{2(-3)^3}{3}-8(-3)\right] \\[2ex] &= \left[\frac{-16}{3}+16\right]-\left[\frac{-54}{3}+24\right] = \frac{38}{3}-8 \\[2ex] &= \frac{14}{3} \end{align} }

${\begin{aligned}\int _{-2}^{2}\left((8-x^{2})-(x^{2})\right)dx&=\int _{-2}^{2}\left(8-2x^{2}\right)dx\\[2ex]&=\left[8x-{\frac {2x^{3}}{3}}\right]{\Bigg |}_{-2}^{2}\\[2ex]&=\left[8(2)-{\frac {2(2)^{3}}{3}}\right]-\left[8(-2)-{\frac {2(-2)^{3}}{3}}\right]\\[2ex]&=\left[16-{\frac {16}{3}}\right]-\left[-16+{\frac {16}{3}}\right]=32-{\frac {32}{3}}\\[2ex]&={\frac {64}{3}}\end{aligned}}$

${\text{Note: }}\int \tan(x)dx=\int {\frac {\sin(x)}{\cos(x)}}dx=\ln |\sec(x)|+C$

${\begin{aligned}u&=\cos(x)\\[2ex]du&=-\sin(x)\\[2ex]-du&=\sin(x)dx\end{aligned}}$

${\begin{aligned}-\int ({\frac {1}{u}})dx&=-\ln |u|+C&=\ln |\cos(x)^{-1}|+C&=\ln |{\frac {1}{cos(x)}}|+C&=\ln |\sec(x)|+C\end{aligned}}$