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

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   \end{cases}
   \end{cases}


<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]


<\math>
<\math>

Revision as of 18:56, 26 August 2022

<math>

\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}

\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>