5.3 The Fundamental Theorem of Calculus/41: Difference between revisions
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\begin{align} | \begin{align} | ||
= \int_{0}^{\frac{\pi}{2}}f(x)\,dx + \int_{\frac{\pi}{2}}^{\pi}f(x)\,dx = \int_{0}^{\frac{\pi}{2}}\sin(x)\,dx + \int_{\frac{\pi}{2}}^{\pi}\cos(x)\,dx \\[2ex] | = \int_{0}^{\frac{\pi}{2}}f(x)\,dx + \int_{\frac{\pi}{2}}^{\pi}f(x)\,dx &= \int_{0}^{\frac{\pi}{2}}\sin(x)\,dx + \int_{\frac{\pi}{2}}^{\pi}\cos(x)\,dx \\[2ex] | ||
= -\cos(x)\bigg|_{0}^{\frac{\pi}{2}} + \sin(x)\bigg|_{\frac{\pi}{2}}^{\pi} = \left[-\cos\left(\frac{\pi}{2}\right) + \cos(0)\right] + \left[\sin(\pi)-\sin\left(\frac{\pi}{2}\right)\right] \\[2ex] | &= -\cos(x)\bigg|_{0}^{\frac{\pi}{2}} + \sin(x)\bigg|_{\frac{\pi}{2}}^{\pi} = \left[-\cos\left(\frac{\pi}{2}\right) + \cos(0)\right] + \left[\sin(\pi)-\sin\left(\frac{\pi}{2}\right)\right] \\[2ex] | ||
= 0+1+0-1 = 0 | &= 0+1+0-1 = 0 | ||
\end{align} | \end{align} | ||
</math> | </math> | ||
Revision as of 22:04, 6 September 2022
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle \int_{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} }
Failed to parse (Conversion error. Server ("https://en.wikipedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\begin{aligned}=\int _{0}^{\frac {\pi }{2}}f(x)\,dx+\int _{\frac {\pi }{2}}^{\pi }f(x)\,dx&=\int _{0}^{\frac {\pi }{2}}\sin(x)\,dx+\int _{\frac {\pi }{2}}^{\pi }\cos(x)\,dx\\[2ex]&=-\cos(x){\bigg |}_{0}^{\frac {\pi }{2}}+\sin(x){\bigg |}_{\frac {\pi }{2}}^{\pi }=\left[-\cos \left({\frac {\pi }{2}}\right)+\cos(0)\right]+\left[\sin(\pi )-\sin \left({\frac {\pi }{2}}\right)\right]\\[2ex]&=0+1+0-1=0\end{aligned}}}