Use a infinite sum to estimate the average value of the function on the given interval by partioning the interval and evaluating the function at the midpoints of the subintervals. f(x)=x^2-3 on [-3, 7]divided into 5 subintervals

Question

anonymous · Answer

Since 5 is not infinity, it wouldn't be an infinite sum.

anonymous · Answer

Interval size is $$
\frac{b-a}{n}=\frac{7-(-3)}{5} = 2
$$

anonymous · Answer

Our values for $x$, will be $$
x_i^*  = \Delta x(i-1)+\Delta{x}/2 = 2(i-1)+1 = 2i-1
$$

anonymous · Answer

All together: $$
\sum_{i=1}^{5}f(x_i^*)\Delta x =
\sum_{i=1}^{5}f(2i-1)\cdot2 =

\sum_{i=1}^{5}((2i-1)^2-3)\cdot 2
$$

aum · Answer

$$
x_i^*  = -3 + \Delta x(i-1)+\Delta{x}/2 = -3 + 2(i-1) + 1 = -3 + 2i - 2 + 1 = 2i - 4
$$