As distance is the integral of speed, you need to use something like the

trapezium rule to estimate the integral:

D~=h/2[y(0) + 2 y(1) + ... + 2 y(n-1) + y(n)]

here D is in nautical miles, h=0.25 hours and so:

D ~= 0.125 [12 + 2*27 + 2*35 + 2*38 + 2*41 + 2*44 + 20] = 50.25 nautical miles

Or Simpson's rule:

D~=h/3[y(0) + 4 y(1) + 2 y(2) + ... +2 y(n-2) + 4 y(n-1) + y(n)]

D ~= 0.08333 [12 + 4*27 + 2*35 + 4*38 + 2*41 + 4*44 + 20] = 51.67 nautical miles

RonL