Finding the value for "k" in a probability density function

[MooreLikeMike]

Hello,

I'm having a hard time figuring out this question: find the value of k so that f(x) = k/(x+1)^2 for x greater than or equal to 1 and less than or equal to 2 is a probability function.

I know that to be a pdf 1) f(x) must be greater than or equal to 1, and 2) the integral of f(x)dx from "a" to "b" must equal 1.

So to solve the question, I found the integral of k/(x+1)^2 and evaluated it at x =2 then subtracted that for when x =1 and got -6/5

It would be great if someone could tell me where I went wrong so that I can try to move in the right direction.

Thank you!

(the work that I've done so far is attached to this)

 

[Steven G]

At the end you plugged in the limits incorrectly. Try again.

 

[MooreLikeMike]

At the end you plugged in the limits incorrectly. Try again.

I think I see what you're talking about. Did you mean that when I was subtracting the F(1) from F(2), I didn't take into account that I was subtracting a negative number, meaning that I should've added F(1) to F(2) instead? Because I see that's a mistake I made.

 

[Harry_the_cat]

I think I see what you're talking about. Did you mean that when I was subtracting the F(1) from F(2), I didn't take into account that I was subtracting a negative number, meaning that I should've added F(1) to F(2) instead? Because I see that's a mistake I made.

Yes that's your mistake. Fix that and you should get k=6.

 

[HallsofIvy]

I know that to be a pdf 1) f(x) must be greater than or equal to 1

No, f(x) must lie between 0 and 1.

 

[MooreLikeMike]

No, f(x) must lie between 0 and 1.

Yeah, I realized I made that mistake as well. Thanks!

 

[MooreLikeMike]

Yes that's your mistake. Fix that and you should get k=6.

Yep, after I adjusted it, that's what I got. Thanks!