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

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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}


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]


using the rule we get
&=6\arcsin{(x)}\bigg|_{\frac{1}{2}}^{\frac{\sqrt{3}}{2}} \\[2ex]
<math>6sin^-1(x)\bigg|_{1/2}^{\sqrt{3}/2}</math>


<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]


[[5.3 The Fundamental Theorem of Calculus/1|1]]
&=\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/3|3]]
 
[[5.3 The Fundamental Theorem of Calculus/5|5]]
&=\pi
[[5.3 The Fundamental Theorem of Calculus/7|7]]
 
[[5.3 The Fundamental Theorem of Calculus/8|8]]
\end{align}
[[5.3 The Fundamental Theorem of Calculus/9|9]]
</math>
[[5.3 The Fundamental Theorem of Calculus/10|10]]
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Latest revision as of 21:59, 6 September 2022

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