5.3 The Fundamental Theorem of Calculus/8: Difference between revisions

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<math>g(x)=\int_{3}^{x}e^{t^2-t}dt</math><br>
<math> g(x)=\int_{3}^{x}e^{t^2-t}dt </math> <br><br>
<math>\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}</math><br>
<math>\frac{d}{dx}\left[g(x)\right] = \frac{d}{dx}\left[\int_{3}^{x}e^{t^2-t}dt\right]=1\cdot(e^{x^2-x})-0\cdot(e^{3^2-3})=e^{x^2-x}</math> <br><br>
Therefore, <math>g'(x)=e^{x^2-x}</math>
<math>\text{Therefore, } g'(x)=e^{x^2-x}</math>

Latest revision as of 20:14, 6 September 2022