Use the Midpoint Rule to approximate the integral from 7 to 17

(1x-1x^2)dx

with n=3

See http://math.fullerton.edu/mathews/n2003/MidpointRuleMod.html

for how to do this. Your "n" is the "m" at that website.
Why do you bother to have the coefficient "1" in front of x and x^2 ?

To approximate the integral using the Midpoint Rule, we divide the interval [7, 17] into n subintervals of equal width. In this case, n is given as 3, so we will have 3 subintervals of width Δx.

Step 1: Calculate the width of each subinterval.
Δx = (17 - 7) / n = 10 / 3 ≈ 3.333

Step 2: Find the midpoint of each subinterval.
To find the midpoint, we add half of the width (Δx) to the left endpoint of each subinterval.
For the first subinterval: midpoint1 = 7 + (Δx / 2) = 7 + (3.333 / 2) ≈ 8.666
For the second subinterval: midpoint2 = 8.666 + (Δx / 2) = 8.666 + (3.333 / 2) ≈ 10
For the third subinterval: midpoint3 = 10 + (Δx / 2) = 10 + (3.333 / 2) ≈ 11.333

Step 3: Evaluate the function at each midpoint.
Substitute the values of x into the given function and calculate the corresponding values.
For the first subinterval: f(midpoint1) = (1 * 8.666) - (1 * 8.666^2) ≈ -55.220
For the second subinterval: f(midpoint2) = (1 * 10) - (1 * 10^2) = -90
For the third subinterval: f(midpoint3) = (1 * 11.333) - (1 * 11.333^2) ≈ -97.367

Step 4: Sum up the values obtained in Step 3.
Approximate the integral by summing up the values obtained in Step 3 and multiplying by the width of the subintervals.
Approximation = Δx * (f(midpoint1) + f(midpoint2) + f(midpoint3))
≈ 3.333 * (-55.220 + (-90) + (-97.367))