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.


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

Do you have any questions about these steps?

  • $\begingroup$ 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

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