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

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Line 6: Line 6:
f(x) =
f(x) =
   \begin{cases}
   \begin{cases}
     sin(x) & 0 \le x < \frac{\pi}{2} \\
     \sin(x) & 0 \le x < \frac{\pi}{2} \\
     cos(x) & \frac{\pi}{2} \le x \le \pi
     \cos(x) & \frac{\pi}{2} \le x \le \pi
   \end{cases}
   \end{cases}


</math>
</math>


<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)




</math>
</math>

Revision as of 18:59, 26 August 2022

Failed to parse (Conversion error. Server ("https://en.wikipedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \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}}}

Failed to parse (Conversion error. Server ("https://en.wikipedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\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)}