Let f(x) = x2, and compute the riemann sum of f over the interval [5, 7], choosing the representative points to be the midpoints of the subintervals and using the following number of subintervals (n). (round your answers to two decimal places.)
(a) use two subintervals of equal length (n = ) use five subintervals of equal length (n = ) use ten subintervals of equal length (n = 10).

a) 72.5

b) 72.64

c) 72.66

Step-by-step explanation:

(a) 2 sub-intervals of equal length (width = 2/2 = 1) would have their mid point at x = 5.5 and x = 6.5.

So the sum over interval [5,7] is

30.25*1 + 42.35*1 = 72.5

(b) 5 sub-intervals of equal length (width = 2/5 = 0.4) would have their mid point at x = 5.2, 5.6, 6, 6.4, 6.8

So the sum over interval [5,7] is

27.04*0.4 + 31.36*0.4 + 36*0.4 + 40.96*0.4 + 46.25*0.4 = 72.64

(c) 10 sub-intervals of equal length (width = 2/10 = 0.2) would have their mid point at x = 5.1, 5.3, 5.5, 5.7, 5.9, 6.1, 6.3, 6.5, 6.7, 6.9

So the sum over interval [5,7] is

0.2 * (26.01 + 28.09+30.25 + 32.49 + 34.81 + 37.21 + 39.69 + 42.25 +44.89 + 47.61) = 72.66

step-by-step explanation:

∆abc has the points a(1, 7), b(-2, 2), and c(4, 2) as its vertices. formed with the point d(1, 2) as its third vertex, then ∆abd is triangle.