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 \\ | ||
</math> | |||
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\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> | ||
--> | |||
<math>\begin{align} | <math>\begin{align} | ||
Revision as of 20:32, 23 August 2022
Failed to parse (syntax error): {\displaystyle g(x)=\int_{3}^{x}e^{t^2-t}dt \\ }
![{\displaystyle {\begin{aligned}u&={\tfrac {1}{\sqrt {2}}}(x+y)\qquad &x&={\tfrac {1}{\sqrt {2}}}(u+v)\\[0.6ex]v&={\tfrac {1}{\sqrt {2}}}(x-y)\qquad &y&={\tfrac {1}{\sqrt {2}}}(u-v)\end{aligned}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/4e0308a3a70f9d0b044c9dd0541c27a130ddbeac)