In this lesson I’m going to explain how to get the expression to calculate a defined integral using Riemann sums.
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Geometric interpretation of Riemann sums
An integral defined in an interval [a,b] gives us the value of the area enclosed between a function f(x) and the x-axis in an interval [a,b], as long as the function is continuous.
Another way to calculate the area enclosed below a curve would be to divide the area into equal rectangles and add the area of each of the rectangles, although this calculation would be approximate:
If we take one of those rectangles:
The base would be the difference of two values of x and the height would be the value of the function for X=Xi
The area of each rectangle would be obtained by multiplying the base by the height and it would remain:
If we make the rectangles smaller and smaller, the calculation of the area becomes more and more exact.
And if rectangles are made infinitely small and we have infinite rectangles, the infinite sum of those rectangles would be the exact area of the enclosed area below that function and would be equal to the defined integral of that function for an interval [a,b]:
From where we obtain the expression used to solve integrals defined by Riemann sums, in which, as we have seen before, the area of each rectangle would be equal to:
Where the increment value of x for an interval [a,b] will be defined as:
And the value of Xi as:
All this is understood much more clearly with an example, which is what we will see below
Example of how to obtain the expression to calculate an integral by the sums of Riemann
Obtain the expression of the Riemann sums of the following integral:
The function in this case is:
For an interval [0,3], therefore a=0 and b=3.
We define the increment of x for that interval:
Y with this incremental expression of x, we calculate Xi:
Now we get the function for X=Xi, substituting x for Xi in the original function:
We calculate the area of each rectangle, multiplying the expressions obtained from f(Xi) and the increment of x:
And we have left:
So the value of the integral by Riemann sums is:
And to obtain the result of the area, we would have to solve the limit of the sum, which we will see in the next section.
How to solve an integral for the sums of Riemann
We are going to solve the expression that was left in the previous section and therefore we will solve the integral by Riemann’s sums.
We start from the previous expression:
We solve the parenthesis by elevating the cube:
And we multiply both fractions:
What is constant we can take out of the summation. Anything that does not carry i is considered constant, so we take the terms that do not carry i out of the summation:
At this point the sum of i elevated to the curo from i=0 to n is equal to this formula:
It can be demonstrated but it is not the objective of this lesson.
We substitute the summation by its expression, according to the previous formula:
We multiply the parenthesis:
And we simplify terms:
And when we solve the limit we are left:
The result is in square units because we are calculating an area.
To show that the result is correct, I am going to solve the integral defined by Barrow’s rule:
And as it could not be otherwise, the result is the same.
Summary formula to solve Riemann sums
Finally, I leave you here the formulas of the sums from the sum of 1 to the sum of i al cubo (that we have used in the example), from 1=0 to n, that you are going to need to solve integrals with Riemann’s sums: