Consider the integral
∫^{8}_2 (3x^2+3x+1)dx
(a) Find the Riemann sum for this integral using right endpoints and n=3.
R3=
(b) Find the Riemann sum for this same integral, using left endpoints and n=3.
L3=
To find the Riemann sum for the given integral using right endpoints and n=3, we need to divide the interval [2, 8] into three equal subintervals and evaluate the function at the right endpoint of each subinterval.
(a) Find the Riemann sum using right endpoints and n=3 (R3):
Step 1: Determine the width of each subinterval.
The width of each subinterval, Δx, is given by:
Δx = (b - a) / n
where "b" is the upper limit of integration (8) and "a" is the lower limit of integration (2), and "n" is the number of subintervals (3).
Δx = (8 - 2) / 3
= 6 / 3
= 2
Step 2: Determine the right endpoints of each subinterval.
Since we are using right endpoints, we need to add the width of each subinterval to the lower limit of integration.
The right endpoints of the three subintervals are:
x1 = 2 + Δx = 2 + 2 = 4
x2 = 4 + Δx = 4 + 2 = 6
x3 = 6 + Δx = 6 + 2 = 8
Step 3: Evaluate the function at the right endpoint of each subinterval.
For each right endpoint, substitute the value of x into the function (3x^2 + 3x + 1) and evaluate it.
R3 = (3(4)^2 + 3(4) + 1) + (3(6)^2 + 3(6) + 1) + (3(8)^2 + 3(8) + 1)
Simplifying the expression:
R3 = (48 + 12 + 1) + (108 + 18 + 1) + (192 + 24 + 1)
= 61 + 127 + 217
= 405
So, the Riemann sum for the given integral using right endpoints and n=3 is R3 = 405.
(b) To find the Riemann sum for the same integral, using left endpoints and n=3 (L3), we need to evaluate the function at the left endpoint of each subinterval.
Step 1: Determine the width of each subinterval.
We have already determined the width of each subinterval as Δx = 2 in the previous part.
Step 2: Determine the left endpoints of each subinterval.
Since we are using left endpoints, the left endpoints of three subintervals are the same as the right endpoints of the previous part:
x1 = 2, x2 = 4, x3 = 6.
Step 3: Evaluate the function at the left endpoint of each subinterval.
For each left endpoint, substitute the value of x into the function (3x^2 + 3x + 1) and evaluate it.
L3 = (3(2)^2 + 3(2) + 1) + (3(4)^2 + 3(4) + 1) + (3(6)^2 + 3(6) + 1)
Simplifying the expression:
L3 = (12 + 6 + 1) + (48 + 12 + 1) + (108 + 18 + 1)
= 19 + 61 + 127
= 207
So, the Riemann sum for the given integral using left endpoints and n=3 is L3 = 207.