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

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<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
<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


&= \-cos(x)
&= \cos(x)


</math>
</math>

Revision as of 18:47, 26 August 2022

Failed to parse (syntax error): {\displaystyle = \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 &= \cos(x) }