Use the composite Simpson's rule to approximate how far a tricycle traveled from a table with the time and velocity

Question

Use the composite Simpson's rule to find how far the car traveled between 15:00 and 15:40 from the following table (Hora/s = Hour/s)

I did

$$I(f)\approx \frac{(b-a)}{3N}[f_0+4f_1+2f_2+4f_3+f_4] \\ = \frac{(b-a)}{3*5}[24.2+4*35+2*41.3+4*42.8+34.8] = \\ (b-a)[30.18667]$$

$$(40)*30.18667 = 1207.45Km$$

That result is ridiculous, obviously, considering this is a tricycle. What went wrong?

Answer 1

If you use hours as the time unit, your data points are: $$ \left\{(0, 24.2),(\frac 16, 35), (\frac 13 , 41.3),(\frac 12, 42.8), (\frac 23, 34.2) \right\} $$

The composite Simpson rule for $s = \int_0^{\frac 23} v(t) dt$ is

$$ \frac{1/6}{3}\left(24.2 + 4 \times 35 + 2 \times 41.3 + 4 \times 42.8 + 34.2 \right) = 25.1222 \,\textrm{Km} $$

if you have any information on the smoothness of $v(t)$, namely about its forth derivative, you can derive an error bound.