5.3 The Fundamental Theorem of Calculus/41: Difference between revisions

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<math>
<math>
\int_{0}^{\pi}f(x)\,dx
\int_{0}^{\pi}f(x)\,dx
\quad \text{where} \;
\quad \text{where} \;
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     \cos(x) & \frac{\pi}{2} \le x \le \pi
     \cos(x) & \frac{\pi}{2} \le x \le \pi
   \end{cases}
   \end{cases}
</math>


</math>


<math>  
<math>  

Revision as of 22:04, 6 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 \int_{0}^{\pi}f(x)\,dx \quad \text{where} \; f(x) = \begin{cases} \sin(x) & 0 \le x < \frac{\pi}{2} \\ \cos(x) & \frac{\pi}{2} \le x \le \pi \end{cases} }


Failed to parse (Conversion error. Server ("https://en.wikipedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\begin{aligned}=\int _{0}^{\frac {\pi }{2}}f(x)\,dx+\int _{\frac {\pi }{2}}^{\pi }f(x)\,dx&=\int _{0}^{\frac {\pi }{2}}\sin(x)\,dx+\int _{\frac {\pi }{2}}^{\pi }\cos(x)\,dx\\[2ex]&=-\cos(x){\bigg |}_{0}^{\frac {\pi }{2}}+\sin(x){\bigg |}_{\frac {\pi }{2}}^{\pi }=\left[-\cos \left({\frac {\pi }{2}}\right)+\cos(0)\right]+\left[\sin(\pi )-\sin \left({\frac {\pi }{2}}\right)\right]\\[2ex]&=0+1+0-1=0\end{aligned}}}