# Riemann Integral Solved Problems To be on the safe side, we look at the largest and smallest possible (and reasonable) rectangles, obtaining the upper sum and the lower sum: Note that since the function f is bounded, the suprema and infima in the definition always exist finite. The upper and lower sum is shown in the following pictures.

To be on the safe side, we look at the largest and smallest possible (and reasonable) rectangles, obtaining the upper sum and the lower sum: Note that since the function f is bounded, the suprema and infima in the definition always exist finite. The upper and lower sum is shown in the following pictures.

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Since the area is always between the upper and lower sum, equality between the two sums means that we determined A.

This manipulation has the form of taking narrower rectangles.

So if we decide to use a different variable in the same formula, the shape and therefore the integral stay the same.

Thus, for instance, Indeed, the area under the same piece of the given parabola is always the same, regardless of what letter we write next to the horizontal axis.

We will do it here, too, but in important statements we will try to write things properly.

For a more thorough explanation of the meaning of the integral notation (not mathematically correct, but very useful for understanding the concept), click here.The error of approximation is then smaller, which means that the upper sum gets smaller (and therefore closer to A) and the lower sum gets larger (and closer to A).In the next picture, compare the error of approximation of the upper and lower sum when we refine the partition.Precisely: We usually just say Riemann integral, it is understood that we mean the definite integral.Since for Riemann integrable functions, the infimum of upper sums is equal to the supremum of lower sums, we could also use the latter to determine the Riemann integral.We also indicated expressions connected with calculating the area of the third rectangle.If we denote the area under the graph of f by A (hoping that it makes sense), then from the picture it seems clear that To determine the area A we will try to manipulate the rectangles so that the upper sum gets smaller and the lower sum gets larger, until they get almost equal.takes on both positive and negative values, then the Riemann sum is the sum of the areas of the rectangles that lie above the x-axis minus sum of areas of the rectangles that lie below the x-axis.We use cookies to make interactions with our website easy and meaningful, to better understand the use of our services, and to tailor advertising.The advantage of the upper/lower sum approach is that we do not have to worry about mechanics of this procedure, all the details are hidden in the definition below.Unfortunately, the Riemann approach using rectangles succeeds only if the function f is nice enough, when f is Riemann integrable.

## Comments Riemann Integral Solved Problems

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The equation is solved in closed form by reduction to two Riemann boundary value problems for.article is devoted to the study of a hypercomplex singular integral equation, which contains as special case the characteristic equation and other ones associated with it.…