6.1 Areas Between Curves/22: Difference between revisions
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<math> | <math> | ||
\int_{0}^{1} \left(\sin\left(\frac{x\pi}{2}\right) - x\right)dx | \int_{0}^{1} \left(\sin\left(\frac{x\pi}{2}\right) - x\right)dx = \int_{0}^{1}\left(\sin\left(\frac{\pi}{2}\right)\right)dx - \int_{0}^{1} (x)dx = \frac{2}{\pi} - \frac{1}{2} | ||
</math> | </math> | ||
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<math> | <math> | ||
\begin{align} | \begin{align} | ||
\int_{0}^{1} \left(\sin(\frac{x\pi}{2})\right)dx \\ | \int_{0}^{1} \left(\sin\left(\frac{x\pi}{2}\right)\right)dx \\ | ||
u = \frac{x\pi}{2} \\ | |||
du = \frac{\pi}{2}dx \\ | |||
\frac{2}{\pi}du =dx \\ | |||
\end{align} | \end{align} | ||
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<math> | <math> | ||
\int_{0}^{1} \left(\sin(\frac{x\pi}{2})\right)dx &= \int_{0}^{\frac{\pi}{2}} \sin(u)du | \begin{align} | ||
\int_{0}^{1} \left(\sin\left(\frac{x\pi}{2}\right)\right) dx &= \frac{2}{\pi} \int_{0}^{\frac{\pi}{2}} \sin(u) du \\ | |||
&= \frac{2}{\pi} \left[-\cos(u)\right]\Bigg|_{0}^{\frac{\pi}{2}} \\ | |||
&= \frac{2}{\pi} \left[-\cos(\frac{\pi}{2})+\cos(0)\right] \\ | |||
&= \frac{2}{\pi} [0+1] = \frac{2}{\pi} \\ | |||
\end{align} | |||
</math> | |||
<math> | |||
\begin{align} | |||
\int_{0}^{1} x dx &= \left[\frac{x^2}{2}\right]\Bigg|_{0}^{1} \\ | |||
&= \frac{1}{2} - 0 = \frac{1}{2} \\ | |||
\end{align} | |||
</math> | </math> | ||
Latest revision as of 03:29, 20 September 2022
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= \sin(\frac{\pi x}{2})} & \color{royalblue}\mathbf{y=x} \\ \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} \sin(\frac{x\pi}{2}) &= x \\ x &= 0 \\ x &=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 \int_{0}^{1} \left(\sin\left(\frac{x\pi}{2}\right) - x\right)dx = \int_{0}^{1}\left(\sin\left(\frac{\pi}{2}\right)\right)dx - \int_{0}^{1} (x)dx = \frac{2}{\pi} - \frac{1}{2} }
Failed to parse (Conversion error. Server ("https://en.wikipedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\begin{aligned}\int _{0}^{1}\left(\sin \left({\frac {x\pi }{2}}\right)\right)dx\\u={\frac {x\pi }{2}}\\du={\frac {\pi }{2}}dx\\{\frac {2}{\pi }}du=dx\\\end{aligned}}}
New upper limit: Failed to parse (Conversion error. Server ("https://en.wikipedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\frac {(0)\pi }{2}}=0}
New lower limit: 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 \frac{(1)\pi}{2} = \frac{\pi}{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_{0}^{1} \left(\sin\left(\frac{x\pi}{2}\right)\right) dx &= \frac{2}{\pi} \int_{0}^{\frac{\pi}{2}} \sin(u) du \\ &= \frac{2}{\pi} \left[-\cos(u)\right]\Bigg|_{0}^{\frac{\pi}{2}} \\ &= \frac{2}{\pi} \left[-\cos(\frac{\pi}{2})+\cos(0)\right] \\ &= \frac{2}{\pi} [0+1] = \frac{2}{\pi} \\ \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}^{1} x dx &= \left[\frac{x^2}{2}\right]\Bigg|_{0}^{1} \\ &= \frac{1}{2} - 0 = \frac{1}{2} \\ \end{align} }