0
$\begingroup$

A probability density function is given by

$$f(x)=\left\{\begin{matrix} ke^{-2x} &x\geq0 \\ 0&otherwise \end{matrix}\right.$$

Find $k$

My attempt,

$k\int_{0}^{\infty }e^{-2x}dx=1$

But I don't know how to solve this integral because of the infinity.

$\endgroup$
3
$\begingroup$

$$k \int_{0}^{\infty}e^{-2x}dx = k \left[\frac{e^{-2x}}{-2} \right]_{0}^{\infty} = k\left[-\frac{1}{e^{\infty}} + \frac{e^{0}}{2} \right] = \frac{k}{2}$$

Do you have any questions about these steps?

$\endgroup$
  • $\begingroup$ How $-\frac{1}{2e^{2x}}$ becomes $\frac{1}{e^{\infty}}$ ? $\endgroup$ – Mathxx Nov 4 '15 at 17:33
  • $\begingroup$ So, we are here: $$\left[ -\frac{1}{2e^{2x}} \right]_{0}^{\infty}$$ We simply plug in the top ($\infty$) and the bottom ($0$). If we plug in infinity, we get: $-\frac{1}{2e^{2\infty}}$ However, $2\infty = \infty$, $e^{\infty} = \infty$, $-\frac{1}{2\infty}=-\frac{1}{\infty}=0.$ To be a bit more precise, I should have more limits here, but this is a good way to show what's going on in this problem. $\endgroup$ – user43395 Nov 4 '15 at 17:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.