5.3 The Fundamental Theorem of Calculus/37: Difference between revisions

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FTC # 2- the <math>\frac{d}{dx}\left(\int\limits_{a(x)}^{b(x)}F(x)dx\right)</math> is F(b)-F(a) where F is the antiderivitive of f such that <math>F^\prime=f</math>
<math>
\begin{align}


37) g(x)=<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]


<math>g^\prime(x)=\frac{d}{dx}\left(\int\limits_{1/2}^{\sqrt{3}/2}\frac{6}{\sqrt{1-t^2}} dt\right)=6sin^{-1}(x)\bigg|_{1/2}^{\sqrt{3}/2}</math>
&=6\arcsin{(x)}\bigg|_{\frac{1}{2}}^{\frac{\sqrt{3}}{2}} \\[2ex]


=<math>6sin^{-1}(\sqrt{3})/2)-(6sin^{-1}(1/2))</math>
&=\left[6\arcsin\left(\frac{\sqrt{3}}{2}\right)\right]-\left[6\arcsin{\left(\frac{1}{2}\right)}\right] \\[2ex]


=<math>\pi</math>
&=\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/1|1]]
&=\pi
[[5.3 The Fundamental Theorem of Calculus/3|3]]
 
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\end{align}
[[5.3 The Fundamental Theorem of Calculus/7|7]]
</math>
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Latest revision as of 21:59, 6 September 2022

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