Worked examples: Definite integral properties 1 | AP Calculus AB | Khan Academy

We want to evaluate the definite integral from 3 to 3 of f of x dx. We're given the graph of f of x and of y equals f of x, and the area between f of x and the x-axis over different intervals.

Well, when you look at this, you actually don't even have to look at this graph over here because, in general, if I have the definite integral of any function f of x dx from, let's say, a to the same value, from one value to the same value, this is always going to be equal to zero. We're going from three to three; we could be going from negative pi to negative pi. It's always going to be zero. One way to think about it is we're starting and stopping here at three, so we're not capturing any area.

Let's do another one. Here, we want to find the definite integral from 7 to 4 of f of x dx. So we want to go from 7 to 4. You might be tempted to say, "Okay, well, look, the area between f of x and the x-axis is 2, so maybe this thing is 2." But the key realization is this area only applies when you have the lower bound as the lower bound and the higher value as the higher bound.

So, the integral from 4 to 7 of f of x dx—this thing, this thing is equal to 2. This thing is depicting that area right over there. So what about this where we've switched it? Instead of going from four to seven, we're going from seven to four. The key realization is if you switch the bounds—and this is a key definite integral property—this is going to give you the negative value.

So, this is going to be equal to the negative of the integral from 4 to 7 of f of x dx. Now, that is this area; f of x is above the x-axis; it's a positive area. So, this thing right over here is going to evaluate to positive 2, but we have that negative out front. So, our original expression would evaluate to negative 2.