00:02
All right, this problem wants us to find the average value of the given function over the interval 3 -4.
00:09
So in order to do that, we just need to do a definite integral.
00:14
So we'll have the average value of f is the integral from 3 to 4 of our function.
00:32
And this will be divided by the interval length.
00:36
But what's really nice about this is the interval length is only one.
00:39
So really we just need to focus on evaluating the integral.
00:44
Okay.
00:45
And in order for us to evaluate that integral, we're going to need to do a u substitution.
00:50
So we're going to say let u equal the exponent of our exponential function.
00:54
So let u equal negative t squared.
00:58
Take the derivative of each side, or i should say take the derivative with respect to t.
01:04
Either way, you get this negative 2t d t.
01:07
And i like to solve for the differential d t so what you should get here is d u over negative 2t and one other thing i would like to do for this scratch work is change the bounds from t bounds into u bounds so we take the values three and four plug them into our formula for u so if we have u of four that should become negative 16.
01:39
If we have u of 3, then that becomes negative 9.
01:44
All right, so what does our average value become? equals...