6.1 Areas Between Curves/15: Difference between revisions

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<math>
<math>
\begin{align}
\begin{align}
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</math>
</math>


<math>\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_{-2}^{2} \left((8-x^2) - (x^2)\right)dx  = \frac{14}{3} + \frac{64}{3} + \frac{14}{3} = \frac{92}{3}</math>
<math>\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  = 2-\ln(2)-1-1-\ln(2)+2 = -2\ln(2)-2+4 = -2\ln(2)+2</math>




<math>
<math>
\begin{align}
\begin{align}
\int_{-3}^{-2}\left((x^2)-(8-x^2)\right)dx &= \int_{-3}^{-2}\left(2x^2-8)\right)dx \\[2ex]
\int_{-\frac{\pi}{3}}^{0}\left[(\tan(x)) - (2\sin(x))\right]dx \\[2ex]
 
&= \left[\ln|\sec(x)|+2\cos(x)\right]\Bigg|_{-\frac{\pi}{3}}^{0} \\[2ex]
 
&= \left[\ln|\sec(0)|+2\cos(0)\right]-\left[\ln|\sec(-\frac{\pi}{3})+2\cos(-\frac{\pi}{3})|\right] \\[2ex]
 
&= \left[0+2\right]-\left[\ln(2)-2(\frac{1}{2})\right] = 2-\ln(2)-1 \\[2ex]
 
&= 2-\ln(2)-1


&= \left[\frac{2x^3}{3}-8x\right]\Bigg|_{-3}^{-2} \\[2ex]
\end{align}
</math>


&= \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]
<math>
\begin{align}


&= \frac{14}{3}
\int_{0}^{\frac{\pi}{3}} \left[(2\sin(x)) - (\tan(x))\right]dx \\[2ex]
&= \left[-2\cos(x)-\ln|\sec(x)|\right]\Bigg|_{0}^{\frac{\pi}{3}} \\[2ex]
&= \left[-2\cos(\frac{\pi}{3})-\ln|\sec(\frac{\pi}{3})|\right] + \left[2\cos(0)+\ln|\sec(0)|\right] \\[2ex]
&= \left[(-2)(1/2)-\ln(2)\right]+\left[2+0\right] = -1-\ln(2)+2 \\[2ex]
&= -1-\ln(2)+2


\end{align}
\end{align}
</math>
</math>


<math>\text{Note: } \int\tan(x)dx=\int\frac{\sin(x)}{\cos(x)}dx=\ln|\sec(x)|+C</math>
<math>
\begin{align}
u &= \cos(x) \\[2ex]
du &= -\sin(x) \\[2ex]
-du &= \sin(x)dx
\end{align}
</math>


<math>
<math>
\begin{align}
\begin{align}


\int_{-2}^{2} \left((8-x^2) - (x^2)\right)dx &= \int_{-2}^{2}\left(8-2x^2\right)dx \\[2ex]
-\int (\frac{1}{u})dx  
&= \left[8x-\frac{2x^3}{3}\right]\Bigg|_{-2}^{2} \\[2ex]
&= -\ln|u|+C
&= \left[8(2)-\frac{2(2)^3}{3}\right] - \left[8(-2)-\frac{2(-2)^3}{3}\right] \\[2ex]
&=\ln|\cos(x)^{-1}|+C
&= \left[16-\frac{16}{3}\right]-\left[-16+\frac{16}{3}\right] = 32-\frac{32}{3} \\[2ex]
&=\ln|\frac{1}{cos(x)}|+C
&= \frac{64}{3}
&=\ln|\sec(x)|+C


\end{align}
\end{align}
</math>
</math>

Latest revision as of 23:34, 28 September 2022

Screen Shot 2022-09-28 at 4.31.18 PM.png

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

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 }

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} \tan(x) &= 2\sin(x) \\ \tan(x)-2\sin(x) &= 0 \\ x &= 0 \\ \end{align} }

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 = 2-\ln(2)-1-1-\ln(2)+2 = -2\ln(2)-2+4 = -2\ln(2)+2}


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_{-\frac{\pi}{3}}^{0}\left[(\tan(x)) - (2\sin(x))\right]dx \\[2ex] &= \left[\ln|\sec(x)|+2\cos(x)\right]\Bigg|_{-\frac{\pi}{3}}^{0} \\[2ex] &= \left[\ln|\sec(0)|+2\cos(0)\right]-\left[\ln|\sec(-\frac{\pi}{3})+2\cos(-\frac{\pi}{3})|\right] \\[2ex] &= \left[0+2\right]-\left[\ln(2)-2(\frac{1}{2})\right] = 2-\ln(2)-1 \\[2ex] &= 2-\ln(2)-1 \end{align} }


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_{0}^{\frac{\pi}{3}} \left[(2\sin(x)) - (\tan(x))\right]dx \\[2ex] &= \left[-2\cos(x)-\ln|\sec(x)|\right]\Bigg|_{0}^{\frac{\pi}{3}} \\[2ex] &= \left[-2\cos(\frac{\pi}{3})-\ln|\sec(\frac{\pi}{3})|\right] + \left[2\cos(0)+\ln|\sec(0)|\right] \\[2ex] &= \left[(-2)(1/2)-\ln(2)\right]+\left[2+0\right] = -1-\ln(2)+2 \\[2ex] &= -1-\ln(2)+2 \end{align} }

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

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