Riemann Sums: Formula & Concept

Instructor: Glenda Boozer
When we want to find the area of the irregular space between the x-axis and a continuous graph, we can't just draw a rectangle in there and take the area; it doesn't fit. If we draw lots of little rectangles, though, it comes closer. If we keep using more and more rectangles that are smaller and smaller, we'll keep getting closer and closer to the true area.

Riemann Sums

Let's go through a little background to lead us along to the formula for Riemann sums. Let's look at any old continuous function (no gaps), from a to b on a graph, like the picture below. The a point will be the lowest point in the colored area, and b will be the highest point.

Suppose we want to find the area between the x-axis and the graph of the function. We often hear it called the area under the curve, but as you can see, when the curve dips below the x-axis, the area is above the curve.

This could be the pattern for an artsy deck in my back yard, or it could be the area of a plot of land bounded by a river. Surveying has been an important skill since people have owned land. The Greeks came up with something not very different from this method, but they never whipped it into a nice formula like Bernhard Riemann did in the 19th century. All this business aside, the big payoff from Riemann sums is that it leads us to the concept and calculation of integrals, which is an important part of calculus.

Riemann sums illustration

Look at the blue rectangles in the first graph above. The area of those four blue rectangles would be closer to the area under the curve than a wild guess, but not very close. Wouldn't the violet rectangles in the second graph be closer? How about the green ones? The yellow ones?


As you see, we can get closer and closer to the area between the curve and the x-axis by drawing smaller and smaller rectangles that touch the curve. We divide the area up by marking points along the x-axis that will mark off the interval between a and b into n segments. In our picture, n is 4 for the blue rectangles, 8 for the violet ones, 16 for the green ones, and 32 for the yellow ones. The width of each little rectangle will be (b - a)/n, since b - a is the length of the interval we are looking at, and we are chopping it up into n pieces. The height of the rectangle is the value of the function at the upper point of the interval, which would be a + (b - a)/n for the first rectangle, a + 2(b - a)/n for the second rectangle, and so on. When we add them up, we have the formula:

Riemann sums formula

Remember, the

Greek letter sigma

Let's do a simple one. The area under the line f(x) = x is a triangle. If we evaluate it from 0 to 4, we can use the formula for the area of a triangle to get the area of ½BH = ½(4)(4) = 8. Let's pretend that we don't know this, however, and try using Riemann sums. First, we'll split the interval into four subintervals, get their areas and add them up:

Riemann sums: 4 intervals

Since a rectangle's area is simply the base times the height, we have S = 4 + 3 + 2 + 1 = 10. That's too high, but it's in the ball park. What if we split the interval into 8 subintervals?

Riemann sums: 8 intervals

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