 Let's take a look at some of the properties of the definite integral. We'll start by finding the value of the integral from a to a of f of x dx. So it helps to remember the geometric interpretation of the definite integral as the area of a region. So this definite integral corresponds to the area of the region below the graph of y equals f of x above y equals 0 from x equals a to x equals a. And that means the region is this line segment here. But this region has area 0. And so our definite integral also has value 0. And this suggests that the definite integral from a to a of anything is equal to 0. How about combining two definite integrals? Suppose I know that the integral from a to b of f of x dx is m. The integral from b to c is n. What about the integral from a to c? And again, a graph can help us organize our thoughts. So the first definite integral represents the area under the graph of y equals f of x above the x-axis from x equals a to x equals b. Meanwhile, the second definite integral represents the area under y equals f of x above the x-axis from x equals b to x equals c. Since the graph of y equals f of x and the x-axis don't change, we can put them both into the same picture. And we know this first region has area m and this second region has area n. So what about this definite integral? To answer that question, let's consider the geometric meaning. Since the integral represents the area under the graph of y equals f of x above the x-axis from x equals a to x equals c, we should be able to find that by adding together the area from x equals a to x equals b, which corresponds to our first definite integral, together with the area from x equals b to x equals c, which corresponds to our second definite integral. And this leads to another important theorem assuming that the values exist. The integral from a to b, plus the integral from b to c, should be equal to the integral from a to c. So let's put our theorems together. Suppose I know that the integral from a to b is l. What about the integral from b to a? And this is sometimes referred to as reversing the direction of integration. Let's pull in our theorems. So this first theorem tells us that we can continue a definite integral. So what if I take a look at the definite integral from a to b, plus the definite integral from b to a? Our theorem tells us that this is equal to the definite integral from a to a. We can drop out that middle point. But we know something about the definite integral from a to a. It's equal to zero. So that means I know that the sum of these two integrals is zero. Since I know the first integral is l, I can solve for the other integral. And that tells me that changing the direction of integration changes the sign of the definite integral. What if I know the value of a definite integral, and I want to find the value of c times a function? Again, a graph can help us organize our thoughts. This first definite integral is the area of the region that's under the graph of y equals f of x above the x-axis over the interval between a and b. Meanwhile, the definite integral we're trying to evaluate is the area of the region under the graph of y equals c f of x over the same interval. But the graph of y equals c times f of x is going to be c times higher than the graph of y equals f of x. So when we try to find its area, every rectangle in the upper, lower, left, or right sum will also have c times the area. And so that means our definite integral will be c times as big. And this suggests the following. Let c be any constant. Provided it exists, the definite integral of c times a function is equal to c times the definite integral. Let's consider another situation to consider the value of a definite integral. First, we want to find the value of the integral of a constant over the same interval, and then we want to find what happens when we add that constant to our function. So again, we'll look at the geometric nature of the definite integral. This definite integral from a to b of c dx is the area under the graph of y equals c over the x-axis and between x equals a and x equals b. And we see that this is a rectangle with height c and width b minus a, so its area is c times b minus a. Now, the definite integral we're looking for is going to be the area under the graph of y equals f of x plus c above the x-axis and between x equals a and x equals b. Now, the graph of y equals f of x plus c equals f of x shifted vertically c units. And if we consider that graph of y equals f of x, we know the area between that graph and the x-axis over the interval from x equals a to x equals b. So when we raise the graph by c units, we're shifting this entire region upward by c units. So we know the red area already, and we just need to figure out what that additional area is. So we see this vertical shift adds a rectangle with height c and width b minus a. So the area increases by c times b minus a. And so the integral from a to b of f of x plus c is going to be l plus c times b minus a. But l was this original integral from a to b of f of x dx, and c times b minus a is this integral from a to b of c. And so this suggests an important theorem, provided that the integrals exist, the integral of a sum is the sum of the integrals. And we can put our theorems together to find the values of many definite integrals. So for example, suppose I know the integral from 0 to 10 and the integral from 8 to 10, and say I want to find the integral from 0 to 8 of something else. So first I might want to work with these two integrals that have the same integrand just over different but overlapping intervals. So I might use this theorem, which allows me to split the interval of integration. And in this case, because I have the integral between 8 and 10, I might want to split the interval at 8. And because I know some of these values, I can substitute them in. And that leaves this integral from 0 to 8 as the only thing we don't know. So we can solve for that. Now you might object and say, well, that's not the answer to the question. And in fact, this is not the answer to the question. But it is a true statement, and these things are almost always useful. So let's go ahead and write that down. In fact, what we're looking for is the integral from 0 to 8 of 3 times f of x. And we're part of the way there. We at least have the right interval from 0 to 8, but we just don't have the right integrand. Well, at that point, we'll pull in our second theorem. And this allows us to multiply the integrand by any constant we choose. So I can multiply f of x by 7, or negative 3, or pi, or square root of 2. Any number I choose, well, how about choosing 3? So our theorem says that the integral of 3 times the function is going to be 3 times the integral of the function. And that allows us to find the value. Or we could get fancier. As with differentiation, it's convenient to determine the type of thing we have. And so here, our integrand is a sum. So we'll pull in our theorem about the integral of a sum. And this allows us to break apart our integral. And now if we look at the individual pieces, these are constant times function. And so we can use our theorem about what happens for the 8 integral of a constant times a function. And finally, we have a bunch of integrals whose values we know. So we can substitute those in and evaluate our integral.