1. ## Integral. Average Value.

My question is about 4D-2. I see that the first step in the solutions manual is:

$\frac{1}{a} \int^{2a}_{a} \frac{1}{x} \;dx$

Why is there a $\frac{1}{a}$ out front of the integral in that step?

Thanks in advance for any help on this problem...

2. ## Re: Integral. Average Value.

Hey sepoto.

The reason is for the 1/a is because that is the length of the interval. Remember that averages are defined to be average = total / length_of_interval. The total is the integral (since it is summing up everything) and the length is 2a - a = a.