5.3 The Fundamental Theorem of Calculus/37: Difference between revisions
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g(x)=\int_{1 | g(x)=\int_{\frac{1}{2}}^{\frac{\sqrt{3}}{2}}\frac{6}{\sqrt{1-t^2}} dt | ||
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} | 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} | ||
Revision as of 21:37, 6 September 2022
Failed to parse (Conversion error. Server ("https://en.wikipedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle g(x)=\int _{\frac {1}{2}}^{\frac {\sqrt {3}}{2}}{\frac {6}{\sqrt {1-t^{2}}}}dtg^{\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}=6sin^{-1}({\sqrt {3}})/2)-(6sin^{-1}(1/2))=\pi }