Question

Find the average value of the function f(x) = (x + 3)^2 on the interval [0, 1]

Answer 1

integrate first to find area, then multiply that number by (1)/(1-0)

Answer 2

The integral is:
\[\frac{1}{b-a}\int\limits_{\alpha}^{\beta}f(x)dx\]
Where the interval is:
\[[a,b]\]
In this case you have [0,1] so that makes the interval length 1. So, in this case, the average value is the area.
So your integral becomes:
\[\int\limits_0^1 (x+3)^2 dx\]
\[\frac{(x+3)^3}{3}|_0^1=\frac{4^3}{3}-\frac{3^3}{3}=\frac{64}{3}-\frac{27}{3}=\frac{37}{3}\]