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

Revision as of 19:04, 25 August 2022

FTC # 2- the ${\frac {d}{dx}}[\int \limits _{a(x)}^{b(x)}F(x)dx]$ is F(b)-F(a) where F is the antiderivitive of f such that $F^{\prime }=f$

37) $\int \limits _{1/2}^{{\sqrt {3}})/2}{\frac {6}{({\sqrt {1-t^{2}}})}}dt$

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

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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>\int\limits_{1/2}^{(\sqrt{3})/2}\frac{6}{(\sqrt{1-t^2})} dt </math>
+) <math>\int\limits_{1/2}^{\sqrt{3})/2}\frac{6}{(\sqrt{1-t^2})} dt </math>
 using the rule we get
-<math>6sin^-1(x)\bigg|_{1/2}^{\sqrt{3})/2}</math>
+<math>6sin^-1(x)\bigg|_{1/2}^{\sqrt{3}/2}</math>
 [[5.3 The Fundamental Theorem of Calculus/1|1]]