The average value of a function takes the midpoint between the highest point on the function and the lowest point on the function within a specific interval. This midpoint is a horizontal "line" that runs from the lower bound (a) to the upper bound (b). This shape then is a rectangle. The idea here is that when you find the average value of the function, you are finding the area of the rectangle that happens to be the area of the often strangely shaped curve. Got that? Anyway, the formula then for the average value involves the integral, since the integral of a function is the same as the area under the curve of the function. That formula is [tex] \frac{1}{b-a} \int\limits^a_b {f(x)} \, dx [/tex]. Using the formula with our particular function, we have [tex] \frac{1}{4-0} \int\limits^4_0 {(3x^2+1)} \, dx [/tex]. Simplifying a bit we have [tex] \frac{1}{4} \int\limits^4_0 {(3x^2+1)} \, dx [/tex]. Integrating we will get [tex] \frac{1}{4}[ \frac{3x^3}{3}+x] [/tex] from 0 to 4. Simplifying THAT a bit we have [tex] \frac{1}{4}[x^3+x] [/tex] from 0 to 4. Subbing in our values, [tex] \frac{1}{4}[(4^3+4)-0]= \frac{1}{4}(68) [/tex] which is 17.
