5.3 The Fundamental Theorem of Calculus/41: Difference between revisions
Jump to navigation
Jump to search
No edit summary |
No edit summary |
||
Line 14: | Line 14: | ||
<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> = \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) | &= -\cos(x)\\[2ex] | ||
</math> | </math> |
Revision as of 18:52, 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 \\[2ex] &= -\cos(x)\\[2ex] }