Answer
$\text{(a)}$
\begin{align}
Average = 5.28
\end{align}
$\text{(a)}$
\begin{align}
Average = 4.31
\end{align}
$\text{(a)}$
\begin{align}
f_{ave} = 4
\end{align}
$\text{(d) See explanation.}$
Work Step by Step
$\text{The given function is}$
\begin{align}
f(x) = 3x^2
\end{align}
$\text{(a)}$
\begin{align}
Average = \frac{f(0.4)+f(0.8)+f(1.2)+f(1.6)+f(2)}{5} = 5.28
\end{align}
$\text{(b)}$
\begin{align}
Average = \frac{f(0.1)+f(0.2) + f(0.3) +...f(1.9)+f(2)}{5} = 4.31
\end{align}
$\text{(c)}$
\begin{align}
f_{ave} = \frac{1}{2} \int_0^2 3x^2 \ dx = 0.5 \left[ x^3\right]_0^2 = 4
\end{align}
$\text{(d)}$
$\text{The average in part (c) is smaller because it considers the average value of the}$
$\text{function over the interval while average values in parts (a) and (b) consider}$
$\text{certain values over the interval.}$
