g ( x ) = ∫ 3 x e t 2 − t d t {\displaystyle g(x)=\int _{3}^{x}e^{t^{2}-t}dt} d d x [ ∫ 3 x e t 2 − t d t ] = 1 e x 2 − x − 0 e ( 0 ) 2 − 0 = e x 2 − x {\displaystyle {\frac {d}{dx}}\left[\int _{3}^{x}e^{t^{2}-t}dt\right]=1e^{x^{2}-x}-0e^{(0)^{2}-0}=e^{x^{2}-x}} therefore, g ′ ( x ) = e x 2 − x {\displaystyle g'(x)=e^{x^{2}-x}}
![{\displaystyle {\frac {d}{dx}}\left[\int _{3}^{x}e^{t^{2}-t}dt\right]=1e^{x^{2}-x}-0e^{(0)^{2}-0}=e^{x^{2}-x}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/4d795c15ec5a644cd7351e7b56328c7f852e9e64)
