 So before proceeding, we'll want to introduce the notation of the definite integral. Now remember that a region is going to need a top, bottom, left, and right. So let's consider the region that's under the graph of y equals f of x, above the graph of y equals g of x, with x equals a as the left side, and x equals b as the right side. Now to commemorate the fact that calculus was invented during the 17th century, we're going to write some of our s's as they did in the 17th century. So this funny looking thing is a 17th century script s. So to approximate the area of the region, we'll partition the region and sum the areas of the rectangles. Again, for convenience, we'll make each of the partitions of equal widths, so we'll make each rectangle delta-wide. And because we'll take this width as a little portion of the x-axis, we'll call this width dx. The other thing we need to do to find the area of a rectangle is to find its height. And because the top of the rectangle is on y equals f of x and the bottom is on y equals g of x, each of our rectangles is going to be f of x minus g of x high. And so the area of any individual rectangle is going to be f of x minus g of x times dx. And that's just one rectangle. And we want to sum all of these rectangles. So we're going to be summing these areas from x equals a to x equals b. And so we designate the area of the region using this notation, which we call the definite integral of f of x minus g of x over the interval a to b. And the good thing to remember is that when we look at the area of the region under the graph of y equals f of x above the graph of y equals g of x and over the interval from a to b, we can express this as the definite integral shown. And again, one of the useful things is to make that transition between the algebraic formulas and expressions and the geometric graphs and pictures. For example, suppose we want to write down the notation for finding the area of the region shown. One way we can do this is to begin by sketching a representative rectangle. So we might sketch something like this. And we want to find the area of this rectangle. Well, that's going to be height times width. And the height is going to be top minus bottom. So remember that the height corresponds to the y values and the top is on this curve y equals f of x while the bottom is on this curve y equals 3. And so the height is going to be f of x minus 3. Next, the width is going to be some tiny portion of the x-axis, which we're descending as dx. And so the area of this representative rectangle is going to be height f of x minus 3 times width dx. And we want to sum these areas from x equals negative 1, our left boundary, to x equals 3, our right boundary. And so we can write this as a definite integral as this. So suppose I want to write down the notation for finding the area between the graphs of y equals 4 minus x squared and y equals 3x. Since we can only find the area of a bounded region, we need to identify the top, bottom, left, and right. And since we don't need to impress anyone, let's go ahead and graph the region. And the two curves do intersect in such a way that they define a closed region. So we just need to look for where those intersection points are located. So find the intersection points. Since one curve is y equals 4 minus x squared and the other is y equals 3x, the two curves will intersect when they have the same x and y values. So we'll set our two equations equal to each other and solve, which gives us a quadratic equation, and we can solve this using the quadratic formula. And once we know the x values of the intersection points, we can compute the y values by substituting them into either equation. And now let's draw a representative rectangle. So if we draw the representative rectangle, we see that the top of the rectangle is on y equals 4 minus x squared and the bottom is on y equals 3x. And so the height, as always, is going to be top minus bottom, which gives us a height of 4 minus 3x minus x squared. The width of the rectangle is a little portion of the x-axis, that we'll call dx. And so the area of this representative rectangle is going to be 4 minus 3x minus x squared times dx. So this will be our integrand, and we'll want to sum up all of these areas from x equals negative 4 to x equals 1. And that gives us the definite integral. Now about another problem. Let's express the area of the region between the graph of y equals 2x plus 3 and the graph of y equals x squared over the interval between 0 and 4. And we can very quickly write down the wrong answer by assuming what is top, bottom, left, and right. However, it doesn't do us any good to write down the wrong answer very quickly. So let's see if we can write down the correct answer, even if it takes us a little bit more time. In order to write down the correct answer, we should graph the region. So now we have to identify top, bottom, left, and right. And the thing to notice here is that the top and bottom curves change at the intersection point, which means we need to find where this intersection point is. And in fact, these graphs intersect at two places, x equals 3 or x equals negative 1. Since x equals negative 1 is outside of our region, we can ignore it safely. So for the region between x equals 0 and x equals 3, the heights of the rectangles are going to be top minus bottom, where the top is the line y equals 2x plus 3, and the bottom is the curve y equals x squared. And so this will be my height of the representative rectangle. The width will be a tiny portion of the x-axis, and I'll sum those areas up from x equals 0 to x equals 3. For the region between x equals 3 and x equals 4, the heights of the rectangles are still top minus bottom, but in this case the top curve is y equals x squared, and the bottom is 2x plus 3. So our heights are going to be and the areas of the representative rectangle will be this times dx, and we'll sum those rectangles up from x equals 3 to x equals 4.