Integration V: Definite Integrals and Negative Area

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Lesson Summary: 
Here we describe how a definite integral and the area it represents can be negative. There are typically two ways this can happen. The first is when the function f(x) is negative. The second is when we integrate in the opposite direction so that ∫baf(x)dx=−∫abf(x)dx.
Lesson Inputs: 
Integration III: Definite Integrals As Upper and Lower Sums
Lesson Outputs: 
Properties of Integrals

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Lesson: 

It is important to note from the very beginning that area cannot be a negative quantity. That doesn't make sense. How can we buy a piece of property that has a negative area? But we have motivated the idea of a definite integral as the sum of the area of many rectangles both here and here. However, what we did in those lessons did not depend upon the idea of area. We merely took some approximate value of a function f(x) within a close interval and then multiplied it by the width of that closed interval. If the function f(x) was negative or if we decided to make the width negative then we could end up with a definite integral whose value is less than zero. So with this in mind we say define "negative area" as nothing more than a definte integral that is less than zero.

First, recall that

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If the height is negative and the width is positive then the area will be negative. Similarly, if the width is negative and the height is positive then the area will be negative.

Since the height of the rectangles in a definite integral is the function f(x), then when the function is negative we can have a definite integral be negative. Recall that

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where <Sign in to see all the formulas> and <Sign in to see all the formulas>. The mi's represent the minimum value of the function in each rectangle and the Mi's represent maximum value of the function in each rectangle. If the function is negative then both mi and Mi will be negative and therefore

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Negative Definite Integral From A Negative Function

Now let's deal with the other way a definite integral can be negative. Recall that when we calculate the value of upper and lower sums (or Riemann sums) we have to take the step of calculating the width of each closed interval in the partition so that <Sign in to see all the formulas>. But what if we did <Sign in to see all the formulas> instead of <Sign in to see all the formulas> when calculating the upper and lower sums? In this case we would have

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and

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Since the beginning and endpoints of each closed interval are reversed isn't this like integrating from b to a rather than from a to b? Unlike many things in Calculus where we logically prove things, the argument above is not a proof. What we are doing is defining

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We would say that we are "integrating in the opposite direction" when we integrate from b to a rather than from a to b. How <Sign in to see all the formulas> and <Sign in to see all the formulas> are related is just by a difference in sign. In other words, if one of them is positive then the other will be negative.

Negative Definite Integral From Integrating In The Opposite Direction

In general we don't know if the result of a definite integral will be positive or negative. In the picture below we can't tell visually if <Sign in to see all the formulas> will be positive or negative. In this lesson we have only shown that it is possible for it to be negative. Furthermore, no matter what the value of <Sign in to see all the formulas> is, we automatically know that <Sign in to see all the formulas>.

Definite Integral Can Be Positive Or Negative