The formula for the average value of a function is [tex] \frac{1}{b-a} \int\limits^b_a {f(x)} \, dx [/tex] where b is the upper bound and a is the lower. For us, this formula will be filled in accordingly. [tex] \frac{1}{2} \int\limits^2_0 {(2x^2+3)} \, dx [/tex]. We will integrate that now: [tex] \frac{1}{2}[ \frac{2x^3}{3}+3x] [/tex] from 0 to 2. Filling in our upper and lower bounds we have [tex] \frac{1}{2}[( \frac{2(2^3)}{3}+3(2))-0] [/tex] which simplifies to [tex] \frac{1}{2}( \frac{16}{3}+6) [/tex] and [tex] \frac{1}{2}( \frac{16}{3}+ \frac{18}{3})= \frac{1}{2}( \frac{34}{3}) [/tex] which is 17/3 or 5.667
