Use the midpoint rule to approximate the integral 6x-9x^2 from 2 to 11 with n=3
Not sure about the terminology of "midpoint rule" in this context, but I will take a stab at it.
the domain is from 2 to 11 or 9 units
you are cutting these into 3 parts, so each width = 3 units
so the critical x values are 2,5,8, and 11, with 2 and 11 as the endvalues
so the midpoints of your three bases are:
3.5, 6.5 and 9.5
f(3.5) = 6(3.5) - 9(3.5)^2 = -89.25
f(6.5) = -341.25
f(9.5) = -755.25
so integral (area)
= 3(-89.25-341.25-755.25) = -3557.25
the answer is negative since the curve is below the x-axis from 2 to 11
so the appr. area is 3557.25
actual area
= ∫(9x^2 - 6x) dx from 2 to 11
= 3x^3 - 3x^2 | from 2 to 11
= 3(11)^3- 3(11)^2 - (3(2)^3 - 3(2)^2 )
= 3993 - 363 - 24 + 12
= 3618