g ( x ) = ∫ 3 x e t 2 − t d t d d x [ g ( x ) ] = d d x [ ∫ 3 x e t 2 − t d t ] = 1 e x 2 − x − 0 e 3 2 − 3 = e x 2 − x Therefore, g ′ ( x ) = e x 2 − x {\displaystyle {\begin{aligned}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{aligned}}}
![{\displaystyle {\begin{aligned}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{aligned}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/f5c5e087afeb8eb42452085889d43d97ce1eee26)