Definite Integrals: the Geometric Meaning of Definite Integrals (Part I)

Definite Integrals (Part I)

The Riemann Sum is a way of approximating the value of a function by adding its values along partitions of the domain. It is named after German mathematician Bernhard Riemann, who published it in 1851. The Riemann sum can be used both to find an exact value and as a numerical method for differential equations.

The definite integral can be defined in the following way. Given a function f(x) defined over a closed interval [a, b], and suppose I is the limit of the Riemann sums, then I is the definite integral of f over [a, b].

For any positive number epsilon, there exists a positive delta such that for any partition P of the interval [a, b] with the division points x0, x1, x2, ... up to xn of norm less than delta and any arbitrary choice of points c_k in each sub-interval, the difference between the Riemann sum and the number I is less than epsilon.

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The notation I is called the integral from a to b of f(x) with respect to x.

The integral sign, here denoted by the letter "F", is defined as the area under a curve from a to b. In this case, "F" is called the integrand, "a" is called the lower limit of integration, and "b" is called the upper limit of integration.

One can replace x with y, u or any other symbol. However, this is always a number which is defined as the area of the region defined by the function y = f(x).

When computing the area of a region using Reimann sums, we have seen that the widths of the subintervals need not be the same and that the point c_k chosen in each subinterval need not be the midpoint or other specific points.

However, for the computation of Reimann sums, it may be convenient to choose the partition such that all sub-intervals have the same width and the point c_k is chosen as the right-hand endpoint of each subinterval.

Now, in this situation, the subdivision points are x0 = the left-hand endpoint, x1 = a + b - a/n.

This is the width of each subinterval, and each subinterval has the same width. That is b - a over n, where b is the beginning of the subinterval and a is its end. Thus, x2 is the next point.

You add the width of the subinterval to x1 and so on. In this case, the width of all these sub-intervals is the same, which equals b - a/n.

Thus, we choose the point c_k to be the right-hand endpoint. This gives us the Riemann sum in this particular situation.

Furthermore, in this situation the norm of the partition tends to zero is the same as n tends to infinity because when n gets bigger and bigger, the width of the sub-interval tends to zero.

And now, what is the geometric meaning of this definite integral? It is indeed the total signed area of the region. We call this a signed area because you see clearly that the area of this region is positive when we compute a definite integral.

This portion of the graph gives us a negative value because it is below the x-axis and this portion gives us a positive value. Therefore, this is not exactly the total area of this region.

The area under the curve can be approximated by Riemann sums and then considered as a limit when the norm tends to zero. But, would this Riemann sum always convert to a value when the norm tends to zero?

Sometimes yes and sometimes no. The Riemann sum of a function has a limit when the norm of the function tends to zero, and we say that the function is integrable when this happens. This theorem is very useful in studying functions.

The integral test states that if a function is continuous over a closed interval, then the function is integrable. More generally, if f(x) is defined over a closed interval [a, b] and does not have any jumps at finer points than those in [a, b], then f(x) is integrable.

This is a famous example of a function that is defined over the closed interval [0, 1] and has the value one if x is a rational number, and zero if x is an irrational number.

Now, this is a function that is not Riemann integrable. That is to say, the definite integral of this function from zero to one does not exist. However, the proof of this fact is more complicated and will not be given here.