A Riemann sum is an approximation of a region's area, obtained by adding up the areas of multiple simplified slices of the region. It is applied in calculus to formalize the method of exhaustion, used to determine the area of a region. This process yields the integral, which computes the value of the area exactly.
Let us decompose a given closed interval into subintervals by inserting points such that
Such a collection of points is called a partition of , and this partition determines the following closed subintervals:
Let denote the closed subinterval of the partition, and let denote the length of the subinterval. Note that need not be the same for each subinterval.
If is defined on the closed interval and is any point in , then a Riemann sum is defined as
A Riemann sum can be visualized as a division of (approximately) the area under the curve on into adjacent rectangles spanning the interval, where the rectangle has width and height . The area of each rectangle is (height times width). The sum of the areas of all rectangles approximates the actual area under on and is equal to the Riemann sum. Note that could be any point in the subinterval, which means the definition leaves some latitude in choosing the height of each of the rectangles.
For more information on definite integrals, see Definite Integrals.
The Riemann sum of a function is related to the definite integral as follows:
Evaluate the Riemann sum pictured below, which is a left-hand approximation of the area under
As we are using the left endpoints of 4 rectangles on the interval we can first find the value of at each left endpoint in the range:
We also know that the interval is of length 4 and we are splitting it into 4 rectangles of equal width. So the width of each rectangle is 1, or .
We are now left with a sum of the area of each rectangle to find the area of the total blue region in the image above:
which, as we can see from the graph, is a slight overestimate of the area under the curve as the function is decreasing on the interval.
Let be a continuous function bounded from below by the -axis. Find an overestimation and an underestimation of the area of the bounded region on the closed interval using Riemann sums.
For simplicity we divide the interval into subintervals of equal length, . By the extreme value theorem has a minimum and a maximum value on each subinterval, and for the subinterval we will denote them by and respectively. At point we construct a rectangle with height which lies below the curve, and at point we construct a rectangle with height which lies partially above the curve.
To compute the area of each rectangle, we find the product of the corresponding length and height:
(We omit the proof of this statement and for now rely on inspection.)
Let us denote by the sum of the areas of all inner rectangles, and by the sum of the areas of all outer rectangles. Then we obtain the formulas
The area under the curve of the bounded function is larger than the total area of the inner rectangles but smaller than that of the outer rectangles. That is, applying the property of inequalities which states if and , then on the inequality above describing the subinterval, we obtain
making an underestimation and an overestimation.
- If , then and we call this the left-hand Riemann sum approximation.
- If , then and we call this the right-hand Riemann sum approximation.
Evaluate the Riemann sum for on the interval , which uses the left endpoint for each of
- a) 10 equal subintervals
- b) 100 equal subintervals.
We divide the interval into subintervals of equal length:
This partition of includes the following points:
Substituting for and , we obtain
Next, we find the left-hand Riemann sum approximation:
where we have used summation formulas to expand the sums.
Therefore, the answers are as follows:
- a) For ,
- b) For ,
Note that is the actual area under the curve, and by adding more rectangles we would achieve a better approximation.
Evaluate the Riemann sum for on the interval , which uses the right endpoint for each of the infinitely many subintervals.
Let's first divide the -axis into many intervals each of width , so the intervals would be
Then we have subintervals each of width and height:
Thus the area of all the subintervals would be given by the following sum (and we call it the Reimann sum):
Therefore, the area would be the limit as of the above expression, which is equal to
We can express definite integral as a limit of the sum of a certain number of terms. Let be a continuous function in the interval . Divide into equal parts such that the width of each part is . Then
The definite integral of a function in the interval can be defined as
With the help of this formula, we can evaluate some simple definite integrals. The process of finding definite integrals with the use of the above formula is known as definite integral as a limit of a sum.
Consider the "limit of sum" formula defined in the previous section, i.e.
Now that we got the concept in the previous section, we will look for the working rules.
- First of all, express the given series in the form .
- Replace for the integral sign for and for
- The lower and upper limits of the integration are the values of and respectively.
Note that the expression is also of the form because .
Find the sum of the series
First, we will express it in the form
This can be rewritten as