Consider the integral
∫^{8}_2 (4x^2+4x+3)dx
(a) Find the Riemann sum for this integral using right endpoints and n=3.
R_3=
(b) Find the Riemann sum for this same integral, using left endpoints and n=3.
L_3=
for 3 intervals, ∆x = (8-2)/3 = 2, so with f(x) = 4x^2+4x+3
∫[2.8] f(x) dx ≈ ∑[k=1..3] f(xk) ∆x
R_3 = 2(f(4)+f(6)+f(8))
L_3 = 2(f(2)+f(4)+f(6))
now just plug and chug
my interpretation of your ∫^{8}_2 is
the integral from 2 to 8.
did you make a sketch of the parabola?
from 2 to 8 is 6. Since you want 3 divisions, dx = 2
right-ended sum = 2( f(4) + f(6) + f(8) )
assuming you have done the f(x)'s
= 2( 83+171+291) = 1090
left-ended sum = 2( f(2) + f(4) + f(6) )
= 2(27+83+171) = 562
appr area = average of those two = (1090+562)/2 = 826
(actual calculus answer: 810)