find the range of $x$ on which $f$ is decreasing, where $f(x)=\int_0^{x^2-x}e^{t^2-1}dt$ I want to find the range of $x$ on which $f$ is decreasing, where
$$f(x)=\int_0^{x^2-x}e^{t^2-1}dt$$
Let $u=x^2-x$, then $\frac{du}{dx}=2x-1$, then $$f'(x)=\frac{d}{dx}\int_0^{x^2-x}e^{t^2-1}dt=\frac{du}{dx}\frac{d}{du}\int_0^{x^2-x}e^{t^2-1}dt=(2x-1)e^{x^4-2x^3+x^2-1}$$
Since $e^{x^4-2x^3+x^2-1}>0$ for all $x\in \Bbb R$ and $2x-1<0\iff x<\frac{1}{2}$. $f$ is decreasing on $(-\infty,\frac{1}{2})$.
Furthermore, $f$ is increasing on $(\frac{1}{2},\infty)$, $f$ is differentiable at $x=\frac{1}{2}$, and $f'(\frac{1}{2})=0$, $f$ attains its minimum value at $x=\frac{1}{2}$.
Am I right?

Everything is fine ! A little bit more can be said:
$f$ is strictly decreasing on $(-\infty,\frac{1}{2}]$
and
$f$ is strictly increasing on $[\frac{1}{2}, \infty).$
