Worked example: Merging definite integrals over adjacent intervals | AP Calculus AB | Khan Academy

What we have here is a graph of y is equal to f of x, and these numbers are the areas of these shaded regions. These regions are between our curve and the x-axis. What we're going to do in this video is do some examples of evaluating definite integrals using this information and some knowledge of definite integral properties.

So, let's start with an example. Let's say we want to evaluate the definite integral going from negative 4 to negative 2 of f of x dx plus the definite integral going from negative 2 to 0 of f of x dx. Pause this video and see if you can evaluate this entire expression.

So, this first part of our expression, the definite integral from negative 4 to negative 2 of f of x dx, we’re going from x equals negative 4 to x equals negative 2. This would evaluate as this area between our curve and our x-axis, but it would be the negative of that area because our curve is below the x-axis.

We could try to estimate it based on the information they've given us, but they haven't given us exactly that value. But we also need to figure out this right over here. Here, we're going from x equals negative 2 to 0 of f of x dx, so that's going to be this area.

If you're looking at the sum of these two definite integrals, notice the upper bound here is the lower bound here. You're really thinking about this as really going to be the same thing as this is equal to the definite integral going from x equals negative 4 all the way to x equals 0 of f of x dx.

This is indeed one of our integration properties: if our upper bound here is the same as our lower bound here, and we are integrating the same thing, well then you can merge these two definite integrals in this way. This is just going to be this entire area, but because we are below the x-axis and above our curve here, it would be the negative of that area.

So, this is going to be equal to negative seven. Let's do another example. Let's say someone were to walk up to you on the street and say quick, here's a graph. What is the value of the expression that I'm about to write down: the definite integral going from 0 to 4 of f of x dx plus the definite integral going from 4 to 6 of f of x dx? Pause this video and see if you can figure that out.

Well, once again, this first part right over here going from 0 to 4, what would it be? It would be this area, which would be this 5 right over here. But then, we would need to subtract this area because this area is below our x-axis and above our curve. We don't know exactly what this is, but luckily, we also need to take the sum of everything I just showed, plus this right over here, and this we're going from 4 to 6, so it's going to be this area.

Once again, when you look at it this way, you can see that this expression is going to be equivalent to taking the definite integral all the way from 0 to 6 of f of x dx. Once again, even if you didn't see the graph, you would know that because in both cases you're getting the definite integral of f of x dx, and our upper bound here is the same as our lower bound here.

So, once again, we're able to merge the integrals. What is this going to be equal to? Well, we have this area here which is 5, and then we have this area which is 6 that was given to us. Since it's below the x-axis and above our curve, when we evaluate it as a definite integral, it would evaluate as a negative 6.

So, this is going to be 5 plus negative 6, which is equal to negative one, and we're done.