5.3 The Fundamental Theorem of Calculus/8

$g(x)=\int _{3}^{x}e^{t^{2}-t}dt$

${\frac {d}{dx}}\left[g(x)\right]={\frac {d}{dx}}\left[\int _{3}^{x}e^{t^{2}-t}dt\right]=1\cdot (e^{x^{2}-x})-0\cdot (e^{3^{2}-3})=e^{x^{2}-x}$

${\text{Therefore, }}g'(x)=e^{x^{2}-x}$