5.3 The Fundamental Theorem of Calculus/37: Difference between revisions
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<math> | <math> | ||
\begin{align} | |||
\int_{\frac{1}{2}}^{\frac{\sqrt{3}}{2}}\frac{6}{\sqrt{1-t^2}}\,dt &= 6\int_{\frac{1}{2}}^{\frac{\sqrt{3}}{2}}\frac{1}{\sqrt{1-t^2}}\,dt\\[2ex] | |||
&=6\arcsin{(x)}\bigg|_{\frac{1}{2}}^{\frac{\sqrt{3}}{2}} \\[2ex] | |||
= | &=\left[6\arcsin\left(\frac{\sqrt{3}}{2}\right)\right]-\left[6\arcsin{\left(\frac{1}{2}\right)}\right] \\[2ex] | ||
=\pi | &=\left[6\cdot\frac{\pi}{3}\right]-\left[6\cdot\frac{\pi}{6}\right] = 2\pi-\pi \\[2ex] | ||
&=\pi | |||
\end{align} | |||
</math> | </math> | ||
Latest revision as of 21:59, 6 September 2022
![{\displaystyle {\begin{aligned}\int _{\frac {1}{2}}^{\frac {\sqrt {3}}{2}}{\frac {6}{\sqrt {1-t^{2}}}}\,dt&=6\int _{\frac {1}{2}}^{\frac {\sqrt {3}}{2}}{\frac {1}{\sqrt {1-t^{2}}}}\,dt\\[2ex]&=6\arcsin {(x)}{\bigg |}_{\frac {1}{2}}^{\frac {\sqrt {3}}{2}}\\[2ex]&=\left[6\arcsin \left({\frac {\sqrt {3}}{2}}\right)\right]-\left[6\arcsin {\left({\frac {1}{2}}\right)}\right]\\[2ex]&=\left[6\cdot {\frac {\pi }{3}}\right]-\left[6\cdot {\frac {\pi }{6}}\right]=2\pi -\pi \\[2ex]&=\pi \end{aligned}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/687cfb7880dbac82492500ce4b3b613ea2951264)