How to show that $f$ is monotone?

Question

Let $f:\mathbb{R}\rightarrow\mathbb{R}$ be defined by $$f(x) = \int_{e^{x^3 +x}}^{1+e^{x^3+x}}e^{r^2} dr$$ for all $x\in\mathbb{R}$. Prove that $f$ is monotone.

I was thinking about Leibniz rule. Actually my main problem is that how to show that f is continuous...

I am struck at this problem as I am in fear to solve this problem.

Please help me and remove the fear from my mind.....

Thank you.

Answer 1

Note that $h(x):=e^{x^3+x}$ is strictly increasing and positive in $\mathbb{R}$ and $e^{r^2}$ is strictly increasing for $r\geq 0$. Then if $x_1<x_2$ then $$f(x_2)-f(x_1)=\int_{h(x_1)+1}^{h(x_2)+1}e^{r^2}\,dr-\int_{h(x_1)}^{h(x_2)}e^{r^2}\,dr=\int_{h(x_1)}^{h(x_2)}( \underbrace{e^{(r+1)^2}-e^{r^2}}_{>0} )\,dr>0,$$ that is $f$ is strictly increasing in $\mathbb{R}$.

Answer 2

For the first derivative of your integral we get $$e^{x^3+e^{2 \left(x^3+x\right)}+x} \left(e^{2 e^{x^3+x}+1}-1\right) \left(3 x^2+1\right)$$ this derivative is positive.

Answer 3

$f(x)$ is in fact the area of $e^{x^2}$ in interval $[e^{x^3+x},e^{x^3+x}+1]$ of constant length $1$. Since both $e^{x^2}$ and $e^{x^3+x}$ are increasing so is the noted area or $f(x)$