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| FTC # 2- the <math>\frac{d}{dx}[\int\limits_{a(x)}^{b(x)}F(x)dx]</math> is F(b)-F(a) where F is the antiderivitive of f such that <math>F^\prime=f</math>
| | <math> |
| | \begin{align} |
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| 37) <math>\int\limits_{1/2}^{(\sqrt{3})/2}\frac{6}{(\sqrt{1-t^2})} dt </math>
| | \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] |
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| using the rule we get 6sin^-1(x)
| | &=6\arcsin{(x)}\bigg|_{\frac{1}{2}}^{\frac{\sqrt{3}}{2}} \\[2ex] |
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| [[5.3 The Fundamental Theorem of Calculus/1|1]] | | &=\left[6\arcsin\left(\frac{\sqrt{3}}{2}\right)\right]-\left[6\arcsin{\left(\frac{1}{2}\right)}\right] \\[2ex] |
| [[5.3 The Fundamental Theorem of Calculus/3|3]] | | |
| [[5.3 The Fundamental Theorem of Calculus/5|5]]
| | &=\left[6\cdot\frac{\pi}{3}\right]-\left[6\cdot\frac{\pi}{6}\right] = 2\pi-\pi \\[2ex] |
| [[5.3 The Fundamental Theorem of Calculus/7|7]]
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| [[5.3 The Fundamental Theorem of Calculus/8|8]]
| | &=\pi |
| [[5.3 The Fundamental Theorem of Calculus/9|9]]
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| [[5.3 The Fundamental Theorem of Calculus/10|10]]
| | \end{align} |
| [[5.3 The Fundamental Theorem of Calculus/11|11]]
| | </math> |
| [[5.3 The Fundamental Theorem of Calculus/13|13]]
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| [[5.3 The Fundamental Theorem of Calculus/15|15]]
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| [[5.3 The Fundamental Theorem of Calculus/17|17]]
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| [[5.3 The Fundamental Theorem of Calculus/19|19]]
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| [[5.3 The Fundamental Theorem of Calculus/20|20]]
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| [[5.3 The Fundamental Theorem of Calculus/21|21]]
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| [[5.3 The Fundamental Theorem of Calculus/23|23]]
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| [[5.3 The Fundamental Theorem of Calculus/25|25]] | |
| [[5.3 The Fundamental Theorem of Calculus/27|27]]
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| [[5.3 The Fundamental Theorem of Calculus/28|28]]
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| [[5.3 The Fundamental Theorem of Calculus/29|29]]
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| [[5.3 The Fundamental Theorem of Calculus/31|31]]
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| [[5.3 The Fundamental Theorem of Calculus/33|33]] | |
| [[5.3 The Fundamental Theorem of Calculus/35|35]]
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| [[5.3 The Fundamental Theorem of Calculus/37|37]]
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| [[5.3 The Fundamental Theorem of Calculus/39|39]]
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| [[5.3 The Fundamental Theorem of Calculus/41|41]]
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| [[5.3 The Fundamental Theorem of Calculus/53|53]]
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![{\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)