The Riemann sum is a sum of sections whose width is Δx, so we have, in general, Σf(x)Δx. It is pretty much the same deal on how we went from Σ to ∫. So for the slope of a straight line or secant line we used Δy/Δx but for the slope of a tangent line we use dy/dx. We decided that once the Δ of the Δy/Δx was infinitesimally small, we would change the notation from Δy/Δx to dy/dx, to remind us that we are dealing with infinitesimals (also called differentials). The we asked, what happens as Δy/Δx gets smaller and smaller, that is, the endpoints of the secant line get closer and closer to c. If we wanted to estimate the slope at a point c, we drew a line between two points on either side of c that were – Δy/Δx and Δy/Δx from c. The concept was developed by using a secant line, or average slope. When you got to differential calculus the problem was not the slope of a line, but the slope of a curve at some point. When you were in algebra you calculated the slope of a line as Δy/Δx. If you have made it to integral calculus, you must have come through algebra and differential calculus, and if so, you have already seen a change in notation, so I’ll start there. You have two questions here, one of notation, how did we go from Σ to ∫, and the other of how did definite integral come about. Could you get not just an approximation, but the exact area under the curve if you took the left side approximation and added it to the right side approximation and then divided that sum by 2? In other words, would the amount that you overestimated the area under the curve be exactly the same amount that you underestimated the area under the curve? Would the value of the overestimation (in either the left-side approximation, or right-side approximation) be equal to the value of the underestimation? My guess is that if the function has a vertical line of symmetry AND the interval along the x-axis extends an equal distance away from the line of symmetry in both directions (in our example -3 to 3 with the line of symmetry being the y-axis) that the amount of overestimation and underestimation would be the same and when you took the average of the two amounts you would be left with the exact value of the area under the curve. They are used to calculate the areas, volumes, etc of arbitrary shapes for which formulas are not defined.Let's say you have a function y=x^2 and you want to find the area under the curve along the interval -3 to 3 and you're delta x was 6 (or any finite number). Riemann Sumsĭefinite integrals are an important part of calculus. Let’s look at Riemann sums with sigma notations. The more the number of rectangles, the closer it takes us to the actual area. These areas are not accurate, but they help a lot in getting an idea about the actual area. The basic idea behind these sums is to divide the area that is supposed to be calculated into small rectangles and calculate the sum of their areas. These formulations help us define the definite integral.
Riemann sums allow us to calculate the area under the curve for any arbitrary function.
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