0
$\begingroup$

What technique needs to be used to integrate these type of functions?

Consider the function: $f(x)=\displaystyle\int_{0}^{x}e^{-t^2}dt$

Show that for $x\in\mathbb{R}$, $f(x)$ increases on $\mathbb{R}$.

$\endgroup$
1
  • 1
    $\begingroup$ $\exp(-t^2)$ is always positive $\endgroup$ Feb 6, 2020 at 11:33

2 Answers 2

4
$\begingroup$

Suppose $x < y$. Then, we have: $$f(y) = \int_{0}^{y}e^{-t^{2}}dt = \int_{0}^{x}e^{-t^{2}}dt + \overbrace{\int_{x}^{y}e^{-t^{2}}dt}^{\ge 0} \ge \int_{0}^{x}e^{-t^{2}}dt = f(x) $$

$\endgroup$
2
$\begingroup$

The derivative of $f(x)$ is $e^{-x^{2}}$ which is always positive. This well-known integral has no analytic expression.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .