Consider the integral
from 3 to 6 S(2x^2+4x+3)dx
(a) Find the Riemann sum for this integral using right endpoints and n=3.
(b) Find the Riemann sum for this same integral, using left endpoints and n=3.
A Riemann sum is an approximation to the area under a curve between defined limits by subdividing the area into n rectangles each of width h. The height is obtained by evaluating the function. If the height is evaluated at the left edge of each rectangle, it is called the left Riemann sum. Similar procedures apply to the Right and Middle Riemann sums.
See for example:
http://en.wikipedia.org/wiki/Riemann_sum
Notice that there is a typo in the equation for the left Riemann sum, the last term should be f(b-Q) where b is the upper limit, and Q is the width of the rectangles.
In your particular case,
a=3 (lower limit)
b=6 (upper limit)
n=3
h= (6-3)/n = 1
f(x)=2x2 + 4x + 3
left Riemann sum
=(f(3)+f(4)+f(5))*1
right Riemann sum
=(f(4)+f(5)+f(6))*1
Post your results if you want a check.
To find the Riemann sum using right endpoints, we divide the interval from 3 to 6 into 3 equal subintervals, each with a width of (6 - 3) / 3 = 1.
(a) Using right endpoints and n = 3:
Let's label the subintervals as follows:
Interval 1: [3, 4]
Interval 2: [4, 5]
Interval 3: [5, 6]
We evaluate the function S(2x^2 + 4x + 3) at the right endpoint of each subinterval and multiply it by the width of each subinterval:
Riemann sum = (4 - 3) * S(2(4)^2 + 4(4) + 3) + (5 - 4) * S(2(5)^2 + 4(5) + 3) + (6 - 5) * S(2(6)^2 + 4(6) + 3)
Simplifying, we get:
Riemann sum = S(28) + S(73) + S(120)
To find the Riemann sum for a given integral, we need to divide the interval of integration into smaller subintervals, evaluate the function at some point within each subinterval, and then sum up these values.
(a) To find the Riemann sum using right endpoints and n=3, we divide the interval [3, 6] into 3 equal subintervals of width Ξx = (b-a)/n = (6-3)/3 = 1. So, the subintervals become [3, 4], [4, 5], and [5, 6].
Next, we evaluate the function S(2x^2+4x+3) at the right endpoint of each subinterval.
For the first subinterval [3, 4], the right endpoint is x = 4.
For the second subinterval [4, 5], the right endpoint is x = 5.
For the third subinterval [5, 6], the right endpoint is x = 6.
Now, we can evaluate the function at each of these right endpoints:
S(2(4)^2+4(4)+3) = S(32+16+3) = S(51)
S(2(5)^2+4(5)+3) = S(50+20+3) = S(73)
S(2(6)^2+4(6)+3) = S(72+24+3) = S(99)
Finally, we add up these values to get the Riemann sum:
S(51) + S(73) + S(99)
(b) To find the Riemann sum using left endpoints and n=3, we use the same interval [3, 6], but this time we evaluate the function at the left endpoint of each subinterval.
For the first subinterval [3, 4], the left endpoint is x = 3.
For the second subinterval [4, 5], the left endpoint is x = 4.
For the third subinterval [5, 6], the left endpoint is x = 5.
Now, we can evaluate the function at each of these left endpoints:
S(2(3)^2+4(3)+3) = S(18+12+3) = S(33)
S(2(4)^2+4(4)+3) = S(32+16+3) = S(51)
S(2(5)^2+4(5)+3) = S(50+20+3) = S(73)
Finally, we add up these values to get the Riemann sum:
S(33) + S(51) + S(73)