5.2 The Definite Integral 

[STARTS AT 23:47]

LECTURER: So let's find the average value here. What we're doing is we're finding the area and then dividing by our interval. So that should make some sense. So our f average equals 1 over b minus a. In this case, were from 5 to 0, or 0 to 5 times the integral of x plus 1 dx, going from 0 to 5. 

So at a function value of 0, this is at 1. x value of 5, that means we are at six. So we're actually looking for this area here. 

Now, based on the fact this is made up of a triangle and a- we have a triangle here. Blue triangle and a rectangle. The area of that is going to be 17.5. So the average value is 3.5. What that means, I'll do that in a different color, is if I go to a y value of 3.5 and I make a rectangle here of width 5, going from 0 to 5, the area of that rectangle would be equal to the integral from 0 to 5. 

So another way to write that formula we had. f average equals 1 over b minus a integral a to b f of x dx. Another way to write that is that the integral from a to b of f of x dx equals f average times b minus a. So rather than being the lower bound or the upper bound, f average is the exact bound to tell us what the integral is.

[ENDS AT 25:59]