5.3 The Fundamental Theorem of Calculus/8: Difference between revisions
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<math> | <math> | ||
g(x)=\int_{3}^{x}e^{t^2-t}dt \ | g(x)=\int_{3}^{x}e^{t^2-t}dt \break | ||
\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} | ||
</math> | </math> | ||
Revision as of 20:30, 23 August 2022
Failed to parse (unknown function "\break"): {\displaystyle g(x)=\int_{3}^{x}e^{t^2-t}dt \break \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} }