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?

$f(x)=\int_{0}^{x^2-x} e^{t^2-1} dt \implies f'(x)=  (2x-1) e^{(x^2-x)^2-1} >0 ~if~ x>1/2$. Hence $f(x)$ in increasing for $x>1/2$ and decreasing for $x<1/2$. Yes you are right. there is a min at $x=1/2$. This one point does not matter, you may also say that $f(x)$ is increasing in $[1/2,\infty)]$ and decreasing on 4(-\infty, 1/2]$.
Note: whether a function increasing or decreasing is decided by two points (not one). For instance, $x_1>x_2 \leftrightarrows f(x_1) > f(x_2).$ If $f(x)$ is decreasing.
