what is the relationship between the definite integral and the riemann sum?

Question

what is the relationship between the definite integral and the riemann sum of the areas of the rectangles under the curve being integrated

this is caluclas!

Answer 1

The relationship between a definite integral and a riemann sum is that they both find the exact area under the curve.

A riemann sum, as you probably know, is just adding up all the rectangles (infinity rectangles) under a curve. When you make this a definite integral, the limit becomes a integral, the function stays the same, the delta x on the riemann sum becomes a dx in the integral.

Answer 2

Riemann Sums And Definite Integrals

Answer 3

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Question:
What is the relationship between the definite integral and the Riemann sum?

Answer:
In a given interval of a non-negative function of one real variable, the area below the graph can be found by creating rectangles, then adding the rectangular areas as the number of them increases to infinity while the area of each approaches zero. This is called the Riemann sum, and there are requirements to be met for this to equal the area under the curve...see sources.

♪♫ The definite integral is *defined* to be the Riemann sum. ♪♫

The integral of f(x)dx from a to b—the definite integral—is equal to the limit as n >> infinity of the sum, the summation from i = 1 to n, of f(x¡) times ∆x—∑f(x¡)∆x—(which is the height of each little rectangle multiplied by the width of each one) over the function from point a to point b—the Riemann sum. (The summation sign, above, should have 'i=1' below it and 'n' above it....Da** this mathematically challenged word processor!)
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Answer 4

Definite Integral of a given funcion in a given interval [a,b] is the limit of the sum of Riemann when number of subintervals in which we divide the interval, tends to infinite.

In fact, that is the reason, Riemann is regarded as one of the founders of modern integral calculus. He achieved to separate the concept of integral from its geometric basis as was conceived since its inception in the XVII century

Answer 5

i think of you like the relationship between the mandatory as an andiderivative and since the section below a curve. think of you're on the curve of a few function f(x). you start up off at a and pass alongside. you're taking small steps to get from one component on the curve to a component dx to the main appropriate interior the x-direction. for each step, undergo in strategies how a techniques you went interior the y-direction. at last you attain x=b and are status on f(b). What occurs in case you upload up all the y-direction displacements for each step? You get the area between f(b) and f(a), i.e. f(b) - f(a). each and every y-direction displacement may well be approximated (o.k., for small displacements) by potential of the spinoff of f at that component, superior by potential of dx. This quickly leverages the definition of the spinoff because of the fact the slope of the function. So, f(b) - f(a) ~= sum f'(x_i) dx. Taking the decrease because of the fact the step length is going to 0 makes the main appropriate component precisely the mandatory and makes equality carry precisely particularly of roughly. in case you replace f with it fairly is antiderivative, F, and further replace f' with f, then we get F(b) - F(a) ~= sum f(x_i) dx, which may well be quickly interpreted because of the fact the section below the curve f, of which F is the antiderivative. Or, shall we've interpreted the above because of the fact the section below f'. that's a chuffed accident, that the sums may well be interpreted because of the fact the element of rectangles or as vertical displacements. i like to think of of the fundamental theorem of calculus as "the place you're is the sum of the place you have been going".

Answer 6

Not good, I'm afraid. You didn't hear it from me, but word around the water cooler is that the reimann sum has been taking a few pulls off the ole bottle with a few not-so-reputable integers and we both know that ain't good.

Answer 7

here's a question. and im not trying to be cute. but seriously what is all that? what are u talking about? thanks