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] | |||
&= -\cos(x)\\[2ex] | &= -\cos(x)\\[2ex] | ||
Revision as of 18:54, 26 August 2022
Failed to parse (Conversion error. Server ("https://en.wikipedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\begin{aligned}\int \limits _{0}^{\pi }f(x)dx\quad {\text{where}}\;f(x)={\begin{cases}sin(x)&0\leq x<{\frac {\pi }{2}}\\cos(x)&{\frac {\pi }{2}}\leq x\leq \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]&=-\cos(x)\\[2ex]\end{aligned}}}