5.3 The Fundamental Theorem of Calculus/37: Difference between revisions
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<math> | |||
g(x)=\int\limits_{1/2}^{\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} | |||
= | =6sin^{-1}(\sqrt{3})/2)-(6sin^{-1}(1/2)) | ||
= | =\pi | ||
</math> | |||
Revision as of 21:36, 6 September 2022