I'm having trouble with this midpoint problem because I'm not sure if my answer is too high.
The problem is:
Measurements of a lake’s width were taken at 15-foot intervals, as shown:
$x= 0\ \ 15\ \ 30\ \ 45\ \ 60\ \ 75\ \ 90\ \ 105\ \ 120$
$f(x)= 0\ \ 15\ \ 18\ \ 20\ \ 19\ \ 23\ \ 24\ \ 22\ \ 12$
Estimate integral $\int_0^{120}f(x) dx$ with $n = 4$, using Midpoint approximation.
For this question, I ended up with $7200$ but compared to the other approx's I had used (left, right, trapezoidal) this number seems way too high. (my other numbers were $1830, 2190$, and $2010$ respectively). Could someone check me on this and either explain why I'm getting such a high number or let me know if i'm just overthinking my answer.
my work is the integral from $0$ to $120$ of $30(15+45+75+105)$ which equals $7200$.
Your midpoint approximation, with $\Delta x=30$, should be
$$\sum_{k=1}^{4} f(x_k^\star)\cdot 30=30\left(15+20+23+22\right)=2400$$
which is much closer to your other approximations. Looks like you were accidentally using $x$ values instead of $f(x)$ values.