You set up correctly the double integral.
By makin' for the inner integral, we have we just treat as a constant, since we're takin' the derivative with respect to Now, find the integration limits and evaluate the rest.
What does mean ? Area?
Instructions state to find the average value of over the region .
with the region R being a triangle with vertices at (0,0), (0,1) and (1,1).
So the base formula for taking the average is:
Drawing the graph, I find it is a triangle anchored at the origin. I'm approaching it from the dy position so I have points from to and it's entering the region in question at and exiting the region at .
My double integral is as follows:
Here's where I run into trouble because I'm not sure how to use a U sub to my advantage on this problem. If I do so , well I don't have a Y in the function, but obviously I do have a 1, so how would I approach this problem? Would I do so I have:
Or would I do something different?
What's the 1/y doing outside the integral? The quantity A is supposed to be the area of the region over which you are taking the average. Since this is a triangle, with base and height both equal to 1, its area is 1/2, so 1/A should be 2.
You don't need to make a substitution. Think of the integral as . As far as the inner integral is concerned, y is a constant, so you can integrate it as . The rest I'll leave to you.
Wow, yeah I feel kinda embarrassed after missing the fact that with Y as the constant is a lot easier than what I must have been thinking, .
So with the area being 2, here's what I have:
Integrating both terms, I have:
so the constants in front cancel out.
Evaluating them from 0 to 1 I have:
Final answer:
Look good to you guys?