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>
<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>  
<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>\text{Therefore, } g'(x)=e^{x^2-x}</math>
<math>\text{Therefore, } g'(x)=e^{x^2-x}</math>
\end{align}
</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}
\text{Therefore, } g'(x)=e^{x^2-x}
</math>
-->
<math>\begin{align}
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{align}</math>

Revision as of 20:38, 23 August 2022



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