Using the old equation, a squared + b squared = c squared. C would be your hypotenuse
so yes it is a correct right triangle
The formula for the average value of a function is
![\frac{1}{b-a} \int\limits^b_a {f(x)} \, dx](https://tex.z-dn.net/?f=%20%5Cfrac%7B1%7D%7Bb-a%7D%20%20%5Cint%5Climits%5Eb_a%20%7Bf%28x%29%7D%20%5C%2C%20dx%20)
where b is the upper bound and a is the lower. For us, this formula will be filled in accordingly.
![\frac{1}{2} \int\limits^2_0 {(2x^2+3)} \, dx](https://tex.z-dn.net/?f=%20%5Cfrac%7B1%7D%7B2%7D%20%20%5Cint%5Climits%5E2_0%20%7B%282x%5E2%2B3%29%7D%20%5C%2C%20dx%20)
. We will integrate that now:
![\frac{1}{2}[ \frac{2x^3}{3}+3x]](https://tex.z-dn.net/?f=%20%5Cfrac%7B1%7D%7B2%7D%5B%20%5Cfrac%7B2x%5E3%7D%7B3%7D%2B3x%5D%20%20)
from 0 to 2. Filling in our upper and lower bounds we have
![\frac{1}{2}[( \frac{2(2^3)}{3}+3(2))-0]](https://tex.z-dn.net/?f=%20%5Cfrac%7B1%7D%7B2%7D%5B%28%20%5Cfrac%7B2%282%5E3%29%7D%7B3%7D%2B3%282%29%29-0%5D%20%20)
which simplifies to
![\frac{1}{2}( \frac{16}{3}+6)](https://tex.z-dn.net/?f=%20%5Cfrac%7B1%7D%7B2%7D%28%20%5Cfrac%7B16%7D%7B3%7D%2B6%29%20%20)
and
![\frac{1}{2}( \frac{16}{3}+ \frac{18}{3})= \frac{1}{2}( \frac{34}{3})](https://tex.z-dn.net/?f=%20%5Cfrac%7B1%7D%7B2%7D%28%20%5Cfrac%7B16%7D%7B3%7D%2B%20%5Cfrac%7B18%7D%7B3%7D%29%3D%20%5Cfrac%7B1%7D%7B2%7D%28%20%5Cfrac%7B34%7D%7B3%7D%29%20%20%20%20%20)
which is 17/3 or 5.667
It would be 6.70 if you round it to the nearest tenth
Range= biggest number - smallest number
so,
67 - 39 = 28