Continuous joint probability density functions

[das1]

Consider the following joint probability distribution function of (X , Y):

a(x + y^2) {0<=x<=2, 0<=y<=2}
0 otherwise

Calculate the value of the constant a that makes this a legitimate probability distribution. (Round your answer to four decimal places as appropriate.)

And then,
For the distribution above, determine the marginal probability density function of Y and use it to calculate the probability that 0.25 < Y < 0.75. (Round your answer to the fourth decimal place as appropriate.)

 

[Jameson]

What have you tried so far? What properties of joint probability distribution functions do you know? There is one property that was used in a thread you made yesterday which is useful here, just extended to two variables.

 

[das1]

OK so this is what I've done:
I set up the integral from 0 to 2 of (x+y^2)dx dy = 1
I've never done an integral with 2 variables before but I plugged it into this calculator:
Integral Calculator - Symbolab
and it gave me 28/3 ? Set that = 1 and divide and get 3/28?

That's my best guess so far, but I could be way off.

 

[Jameson]

You've never worked with double integrals before? Do you know how to solve something like this?

$$\int_{0}^{1} \int_{1}^{5} x+ydydx$$

If not you haven't been given the tools you need to solve this problem so it's totally understandable why this is difficult! :)

You are on the right track though in your reasoning. The double integral of the joint pdf should equal one:

$$\int_{0}^{2} \int_{0}^{2} a(x+y^2)dydx=1$$