5.3 The Fundamental Theorem of Calculus/37: Difference between revisions
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<math> | <math> | ||
\begin{align} | |||
g(x)=\int_{\frac{1}{2}}^{\frac{\sqrt{3}}{2}}\frac{6}{\sqrt{1-t^2}} dt | g(x)=\int_{\frac{1}{2}}^{\frac{\sqrt{3}}{2}}\frac{6}{\sqrt{1-t^2}} dt \\[2ex] | ||
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 (unknown function "\begin{align}"): {\displaystyle \begin{align} g(x)=\int_{\frac{1}{2}}^{\frac{\sqrt{3}}{2}}\frac{6}{\sqrt{1-t^2}} dt \\[2ex] 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} =6sin^{-1}(\sqrt{3})/2)-(6sin^{-1}(1/2)) =\pi }