5.3 The Fundamental Theorem of Calculus/15: Difference between revisions
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\begin{align} | \begin{align} | ||
y=\int_{0}^{tan(x)}\sqrt{t+\sqrt t}\,dt =\sec^{2}(x) | y=\int_{0}^{tan(x)}\sqrt{t+\sqrt t}\,dt =\sqrt{tan(x)+\sqrt tan(x)}\,\sec^{2}(x) | ||
\end{align} | \end{align} | ||
</math> | </math> | ||
Revision as of 19:22, 25 August 2022
Use part 1 of the FTC to find the derivative of the function: Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle y=\int_{0}^{tan(x)}\sqrt{t+\sqrt t}\,dt}
Failed to parse (Conversion error. Server ("https://en.wikipedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\begin{aligned}y=\int _{0}^{tan(x)}{\sqrt {t+{\sqrt {t}}}}\,dt={\sqrt {tan(x)+{\sqrt {t}}an(x)}}\,\sec ^{2}(x)\end{aligned}}}