# Thread: Complex Coordinate Riemann Sum Calculation & Error

1. ## Complex Coordinate Riemann Sum Calculation & Error

Hello,

I have two calculators on my desk, the first being an HP-50G that I purchased awhile back, and the second, a TI-83+ that is my sister's. Both calculators compute the following problem differently:

You will be working with the function $y = f(x) = (x - 1.5)^{\frac{1}{3}} + 2$, and the interval you will always be concerned about is [-2, 6]. Here's a graph:

The actual area under the curve on the interval [−2, 6] is about 17.586.

Approximating the area under the curve on the interval [−2, 6] by using a Riemann sum with 4 equal subdivisions and left-hand endpoints gives you
R = 2(f(−2) + f(0) + f(2) + f(4)) = 14.976. Draw the rectangles used for this Riemann sum on one of your paper graphs. Do you see where the error for this approximation is?

Approximating the area under the curve on the interval [−2, 6] by using a Riemann sum with 4 equal subdivisions and right-hand endpoints gives you
R = 2(f(0) + f(2) + f(4) + f(6)) = 21.314. Draw the rectangles used for this Riemann sum on one of your paper graphs. Do you see where the error for this approximation is?

The TI-83+ gives the expected 14.976, but the HP-50G gives the complex coordinate (5.357209, 2.35075461244), though it does not list $i$. Why does the TI-83+ list a real number when the HP-50G perceives it as being a complex coordinate?

I also produced another graph with $n = 4$ using left endpoints:

As far as "error" they aren't very explicit; I would venture to say that the error in the approximation is the area sections that are left white (that hasn't been added to the area).

2. ## Re: Complex Coordinate Riemann Sum Calculation & Error

Originally Posted by Biff
You will be working with the function $y = f(x) = (x - 1.5)^{\frac{1}{3}} + 2$, and the interval you will always be concerned about is [-2, 6].
The actual area under the curve on the interval [−2, 6] is about 17.586.

Approximating the area under the curve on the interval [−2, 6] by using a Riemann sum with 4 equal subdivisions and left-hand endpoints gives you
R = 2(f(−2) + f(0) + f(2) + f(4)) = 14.976. Draw the rectangles used for this Riemann sum on one of your paper graphs. Do you see where the error for this approximation is?

Approximating the area under the curve on the interval [−2, 6] by using a Riemann sum with 4 equal subdivisions and right-hand endpoints gives you
R = 2(f(0) + f(2) + f(4) + f(6)) = 21.314. Draw the rectangles used for this Riemann sum on one of your paper graphs. Do you see where the error for this approximation is?

The TI-83+ gives the expected 14.976, but the HP-50G gives the complex coordinate (5.357209, 2.35075461244), though it does not list $i$.
I am not all sure as to what you asking.

Note the first sum is a left hand sum.

The second sum is right hand sum.

The average is close to the actual.