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>
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>


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


using the rule we get
<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>
<math>6sin^{-1}(x)\bigg|_{1/2}^{\sqrt{3}/2}</math>


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

Revision as of 19:30, 25 August 2022

FTC # 2- the is F(b)-F(a) where F is the antiderivitive of f such that

37) g(x)=

=

=

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