Prove that the average of the right Riemann sum and the left Riemann sum, using n rectangles to approximate the area beneath the curve f(x) on the interval [a,b], is equal to the Trapezoidal Rule approximation for the same function on the same interval using n trapezoids.
Difficulty:
Solution:Before we look at a general proof, let me show you a specific example. We'll walk through the plan for this case, and it should make the general case much easier to understand. Below, I have graphed some function f(x) under which we'll approximate the area using left and right Riemann sums with n = 7 rectangles. The blue line is f and the red lines represent the function values at each subinterval.
Let's say we're trying to approximate the area on the interval
. The width of each interval above will be
. It's no hard task to then calculate the left sum, which I will denote
:
Similarly, you can calculate the right sum,
:
In order to find the average of these two sums, we add them and divide by 2. In practical terms, that means to add the terms in parentheses, leave the common factor of
Recognize this puppy? It's the Trapezoidal Rule applied to our function f with n = 7 trapezoids. So, that's the proof in essence. However, that proves it for just this one case--we want a generic proof that works for all cases. No problem. We've already done it; just replace all the 7's above with n's. Therefore, for a continuous function f(x) on n intervals, you get the left and right sums
The average of these sums is, once again, the formula for the Trapezoidal Rule for the function f with n trapezoids:
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