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

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   \end{cases}
   \end{cases}


<\math>
<math>
&= \int\limits_{0}^{\frac{\pi}{2}}f(x)dx + \int\limits_{\frac{\pi}{2}}^{\pi}f(x)dx = \int\limits_{0}^{\frac{\pi}{2}}\sin(x)dx + \int\limits_{\frac{\pi}{2}}^{\pi}\cos(x)dx \\[2ex]
&= \int\limits_{0}^{\frac{\pi}{2}}f(x)dx + \int\limits_{\frac{\pi}{2}}^{\pi}f(x)dx = \int\limits_{0}^{\frac{\pi}{2}}\sin(x)dx + \int\limits_{\frac{\pi}{2}}^{\pi}\cos(x)dx \\[2ex]


&= -\cos(x)\\[2ex]
<\math>
 
<math>
-\cos(x)\\[2ex]


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

Revision as of 18:54, 26 August 2022

Failed to parse (unknown function "\math"): {\displaystyle \begin{align} \int\limits_{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} <\math> <math> &= \int\limits_{0}^{\frac{\pi}{2}}f(x)dx + \int\limits_{\frac{\pi}{2}}^{\pi}f(x)dx = \int\limits_{0}^{\frac{\pi}{2}}\sin(x)dx + \int\limits_{\frac{\pi}{2}}^{\pi}\cos(x)dx \\[2ex] <\math> <math> -\cos(x)\\[2ex] \end{align} }