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>


37) g(x)=<math>\int\limits_{1/2}^{\sqrt{3}/2}\frac{6}{\sqrt{1-t^2}} dt </math>
g(x)=\int\limits_{1/2}^{\sqrt{3}/2}\frac{6}{\sqrt{1-t^2}} dt


<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>
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>6sin^{-1}(\sqrt{3})/2)-(6sin^{-1}(1/2))</math>
=6sin^{-1}(\sqrt{3})/2)-(6sin^{-1}(1/2))


=<math>\pi</math>
=\pi


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</math>
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Revision as of 21:36, 6 September 2022