5.3 The Fundamental Theorem of Calculus/8: Difference between revisions
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<math> | <math> | ||
\begin{align} | \begin{align} | ||
g(x)=\int_{3}^{x}e^{t^2-t}dt | g(x)=\int_{3}^{x}e^{t^2-t}dt | ||
\frac{d}{dx}\left[g(x)\right] = \frac{d}{dx}\left[\int_{3}^{x}e^{t^2-t}dt\right]=1e^{x^2-x}-0e^{3^2-3}=e^{x^2-x} | \frac{d}{dx}\left[g(x)\right] = \frac{d}{dx}\left[\int_{3}^{x}e^{t^2-t}dt\right]=1e^{x^2-x}-0e^{3^2-3}=e^{x^2-x} | ||
\text{Therefore, } g'(x)=e^{x^2-x} | \text{Therefore, } g'(x)=e^{x^2-x} | ||
\end{align} | \end{align} | ||
</math> | </math> | ||
Revision as of 20:28, 23 August 2022
Failed to parse (Conversion error. Server ("https://en.wikipedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\begin{aligned}g(x)=\int _{3}^{x}e^{t^{2}-t}dt{\frac {d}{dx}}\left[g(x)\right]={\frac {d}{dx}}\left[\int _{3}^{x}e^{t^{2}-t}dt\right]=1e^{x^{2}-x}-0e^{3^{2}-3}=e^{x^{2}-x}{\text{Therefore, }}g'(x)=e^{x^{2}-x}\end{aligned}}}