 Because the definite integral is the limit of a Riemann sum, it will give us a signed area. And what this means is you should always identify top, bottom, left, and right. What happens if we don't? Well, let's take an example. Suppose we want to find the area between the graph of y equals x and the x-axis over the interval from negative 5 to 5. Well, since this is a geometry problem, we're finding area, we'll make this as hard as possible and refuse to draw the graph. And, well, let's see, area is the integral, so we'll set up the integral. We start at negative 5, go to 5, the graph is y equals x, and so we have this integral. We'll find the antiderivative, we'll evaluate, and we'll get our area of 0. Let's draw the graph. And the region between is this part. The important thing to notice here is that our graph y equals x is the top sometimes. And we can figure out that where that crossing occurs is at x equal to 0. And so we have two regions. So the correct area is the area of region 1 plus the area of region 2. In the first region, the top is the x-axis and the bottom is y equals x. So if we draw a representative rectangle, we see that these rectangles have a top at y equals 0, a bottom at y equals x, and the height, top minus bottom, 0 minus x, negative x, at the width, a tiny portion of the x-axis, otherwise known as dx, and we'll sum them from x equals negative 5 to x equals 0. In the second region, we see that the top is the line y equals x, and the bottom is y equals 0. And so these rectangles have a top, y equals x, bottom, y equals 0, with dx, and we'll sum them from x equals 0 up to x equals 5. And now we can anti-differentiate, we can evaluate, and the sum will be the area of the two regions put together, which is to say the area of the whole region. Or we might try to find the area between the graph of y equals sine of x and the x-axis over the interval from 0 to 2 pi. And if we graph, we see that the area we want is actually two regions put together. So if we draw our representative rectangles in the first region, our rectangles have top at y equals sine x, the bottom is y equals 0, and the height, top minus bottom, sine x minus 0, sine of x, and with dx. And we notice that this first region extends from x equals 0 up to x equals pi. And then we'll sum these rectangles from x equals 0 up to x equals pi. In our second region, we have our rectangles with top at y equals 0, bottom at y equals sine x, height, top minus bottom, 0 minus sine of x, or negative sine x, with dx, and we'll sum these from x equals pi up to x equals 2 pi. At that point, we have our integrals, we anti-differentiate, we evaluate, and we get our area.