6.1 Areas Between Curves/22: Difference between revisions
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\int_{0}^{1} \left(\sin(\frac{x\pi}{2})\right)dx &= \int_{0}^{\frac{\pi}{2}} \sin(u)du \\ | \int_{0}^{1} \left(\sin(\frac{x\pi}{2})\right)dx &= \int_{0}^{\frac{\pi}{2}} \sin(u)du \\ | ||
& u = \frac{x\pi}{2} \\ | & u = \frac{x\pi}{2} \\ | ||
& du = \frac{\pi}{2}dx \quad b=\frac{(0)\pi}{2}=0 \\ | & du = \frac{\pi}{2}dx \quad b= \frac{(0)\pi}{2}=0 \\ | ||
& \frac{2}{\pi}du=dx \quad a=\frac{(1)\pi}{2} = \frac{\pi}{2} \\ | & \frac{2}{\pi}du=dx \quad a= \frac{(1)\pi}{2} = \frac{\pi}{2} \\ | ||
& b= \frac{(0)\pi}{2} = 0 \\ | & b= \frac{(0)\pi}{2} = 0 \\ | ||
Revision as of 03:02, 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 }
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(\frac{x\pi}{2})\right)dx &= \int_{0}^{\frac{\pi}{2}} \sin(u)du \\ & u = \frac{x\pi}{2} \\ & du = \frac{\pi}{2}dx \quad b= \frac{(0)\pi}{2}=0 \\ & \frac{2}{\pi}du=dx \quad a= \frac{(1)\pi}{2} = \frac{\pi}{2} \\ & b= \frac{(0)\pi}{2} = 0 \\ & a= \frac{(1)\pi}{2} = \frac{\pi}{2} \\ \end{align} }