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.
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.
1 answer
1 answer