5.3 The Fundamental Theorem of Calculus/8

$g(x)=\int _{3}^{x}e^{t^{2}-t}dt$ \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{align} </math>

${\begin{aligned}u&={\tfrac {1}{\sqrt {2}}}(x+y)\qquad &x&={\tfrac {1}{\sqrt {2}}}(u+v)\\[0.6ex]v&={\tfrac {1}{\sqrt {2}}}(x-y)\qquad &y&={\tfrac {1}{\sqrt {2}}}(u-v)\end{aligned}}$