Mark M. answered • 12/03/23
I love tutoring Math.
Let f(x) = x/2 + x5.
Your left and right limits of integration are 3 and 6, an interval of width 3. Let's divide this interval into 10 equal subintervals, each of length .3 (And let's call this .3 "delta x"). The 10 subintervals are
from 3.0 to 3.3
from 3.3 to 3.6
from 3.6 to 3.9
from 3.9 to 4.2
from 4.2 to 4.5
from 4.5 to 4.8
from 4.8 to 5.1
from 5.1 to 5.4
from 5.4 to 5.7
from 5.7 to 6.0
To generate the ten numbers 3.3, 3.6, 3.9, ..., 6.0 that are the "right endpoints" of the ten subintervals, consider the expression
3+(3i)/10.
When i=1, the value of this expression is 3.3
When i=2, the value of this expression is 3.6
When i=3, the value of this expression is 3.9
When i=4, the value of this expression is 4.2
etc.
When i=10, the value of this expression is 6.0
See? 3+(3i)/10 is a little machine that manufactures the series of numbers 3.3, 3.6, 3.9, 4.2, ..., 6.0
(The first "3" in the machine is there because it is the lower limit of integration.
The second "3" in the machine is there because it is the distance between the two limits of integration.
The "10" in the machine is there because it is the number of subintervals.)
If i = 1, then the area of a rectangle whose height is f(3.3) and whose width is 3 would be height times width
= f(3.3) times .3 = f(3+(3i)/10) times delta x
If i = 2, then the area of a rectangle whose height is f(3.6) and whose width is 3 would be height times width
= f(3.6) times .3 = f(3+(3i)/10) times delta x
If i = 3, then the area of a rectangle whose height is f(3.9) and whose width is 3 would be height times width
= f(3.9) times .3 = f(3+(3i)/10) times delta x
etc.
If i = 10, then the area of a rectangle whose height is f(6.0) and whose width is 3 would be height times width
= f(6.0) times .3 = f(3+(3i)/10) times delta x
Therefore the sum of the areas of these 10 rectangles, approximating the area under the curve from 3 to 6,
could be written
∑ (from i=1 to 10) f(3 + (3i)/10) times delta x
where delta x = .3 (one tenth of the distance from 3 to 6).
If we used 100 rectangles instead of just 10, the sum would be
∑ (from i=1 to 100) f(3 + (3i)/100) times delta x
where delta x = .03 (one hundredth of the distance from 3 to 6).
If we used n rectangles instead of just 10, the sum would be
∑ (from i=1 to n) f(3 + (3i)/n) times delta x
where delta x = 3/n (one nth of the distance from 3 to 6).
Take the limit as n goes to infinity:
lim (n→∞) ∑ (from i=1 to n) f(3 + (3i)/n) times delta x
That's the Riemann limit you want. Good luck.
