5.3 The Fundamental Theorem of Calculus/41: Difference between revisions
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\end{cases} | \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> | <\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>