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 (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}{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{align} }