Question

how do i find the integral of (2/3)(x+2y)dy

Answer 1

I would first distribute to get 2/3(x) + 4/3(y) and then integrate. When you integrate with respect to y 2/3(x) is treated like a constant so you get 2/3(xy) + 2/3y^2 + c

Answer 2

yeah that's what i thought...but its not working out right....maybe i just made a simple mistake on there....it doesn't change it that its integrated from 0 to 1 does it

Answer 3

\[\int\limits_{?}^{?}(2/3)(x+2y)dy\] Take out the 2/3 to get \[(2/3)\int\limits_{?}^{?}(x+2y)dy\] Then, since you are taking the integral with respect to y, you treat x like a constant and get: \[(2/3)[xy+(2y ^{2})/2] +c\] Simplify to get:\[(2/3)[xy+y ^{2}]+c\]

Answer 4

And yes! Just as TBates posted, you can distribute that 2/3.

Answer 5

The answer does change if you have bounds from 0 to 1. That means you don't have that +c at the end.

Answer 6

when i completed it I got (2/3)x+(2/3)

Answer 7

If you're integrating over a region (I'm assuming the 0 to 1 is y values) you substitute 1 in for y and subtract off what you get when you substitute 0 in for y and you should get 2/3[x+1], as long as I did my math right.

Answer 8

well it'sactually x values (i think) this is applied statistics and I am finding the marginal densities

Answer 9

so for the marginal density of X it is g(x)= the integral of f(x,y)dy for 0<x<1 and 0<y<1

Answer 10

this is a sample problem that I keep getting a different answer than the book...so I am trying to figure it out so that I can do my actual homework problem

Answer 11

ok, then the answers we provided 2/3x + 2/3 would be correct because if you're looking for marginal for X you sum up all your y's.
On a side note if you take the integral of 2/3x + 2/3 from 0 to 1 you get 1 (the total probability)

Answer 12

oh my goodness...i kept getting that answer but was thinking it was wrong because i was looking in the wrong spot in my book...thank you so much for your help