# Definite Integral prob

1. Dec 3, 2006

### maxpayne_lhp

Hello,

Okay so I was given a density function:

$$f(x) = 2e^(-0.25x)$$

The problem asks for the value of Pr(X < or = 3)

I first figured out the probability density function first by let

$$\int (3,0) k.25e^(-.25x) = 1$$

And figured out that k = .45

and continue solving my problem until i get the final answer is -.212

So, is my solution correct? Did I misunderstand something about k?

Thanks much!

NN

2. Dec 3, 2006

### maxpayne_lhp

Sorry, my LaTeX code looks bad.

3. Dec 4, 2006

### dextercioby

Click this picture to see where your typing went wrong

$$k\int_{0}^{\infty} e^{-0.25 x}{}dx =1$$

I don't know about the value of k, whether it's correct or not, guess not...

Daniel.

4. Dec 4, 2006

### HallsofIvy

Staff Emeritus
What do you mean, you were "given a density function:
$f(x)= 2e^{-.25x}$"? You seem to think that you need to find a multiplier k so that the "total integral" (is that supposed to be an integral from 0 to 3?) is 1. If that were the case then you wouldn't be asked for the probability that x is between 0 and 3: it would be 1!

Dextercioby, on the other hand, seems to think that you need to find k such that the integral from 0 to infinity is 1.

I see that we are already told that the function is $2e^{-.25x}$. That is, that the "k" is 2, but that we are not given an interval over which this is to be the probability density.

5. Dec 4, 2006

### maxpayne_lhp

Daniel, yeah thanks for the code... i meant

$$k\int_{0}^{3} e^{-0.25 x}{}dx =1$$

HallsofIvy, Uh... Thats the same word from the problem in the book... and yes, thats the integral from 0 to 3... Im not sure I know about the total integral....

Then please state the problem exactly, word for word. So far you have told us that the density function is $$f(x) = 2e^{-0.25x}$$
that $$k\int_{0}^{3} e^{-0.25 x}{}dx =1$$