# find the range of $x$ on which $f$ is decreasing, where $f(x)=int_0^{x^2-x}e^{t^2-1}dt$

Mathematics Asked by Steven Lu on December 4, 2020

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 $$xin Bbb R$$ and $$2x-1<0iff x. $$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.

Answered by Z Ahmed on December 4, 2020

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 $$[frac{1}{2}, infty).$$

Answered by Fred on December 4, 2020