 By remembering the geometric meaning of the definite integral, it's possible to gain other information about the definite integral from just a few pieces of data. For example, suppose I know that the definite integral from 0 to 8 is 16, and suppose we also know that g of x is increasing and g of 5 is equal to 2. Let's see if we can find an upper bound for the definite integral between 0 and 5. Now it's possible to solve this problem without drawing anything. And indeed, if you want to impress a rogue mathematician doing a problem like this without drawing anything is a good way to do it. But it's easier to draw some pictures. So if we take the fact that this integral has values 16 and combine it with the geometric meaning of the definite integral, we find that the area of the region below y equals g of x above the x axis from x equals 0 to x equals 8 has area 16. Now the problem is we don't know what g of x looks like, but we do know a few things about it. We know that g of x is increasing, and that tells us that the graph of y equals g of x is rising. So it's got to look like no, no, no, yes. And then finally we're told that g of 5 is equal to 2. And so we know that this graph has to pass through the point 5, 2. Well, the graph we drew while it's increasing doesn't pass through this point, so let's fix it. Now remember the question asks us to say something about the definite integral from 0 to 5 of g of x. Now something we might do is we might write down any relationship between the things we know and the thing we're looking for. And in this case, the thing that we know is this definite integral, and the thing we're looking for is this definite integral. And so we might write down this sum based on the additivity of the definite integral. This suggests that I might want to say something about the definite integral from 5 to 8. So let's consider the geometric content of this definite integral. Now this corresponds to the area under the graph between x equals 5 and x equals 8. Can we say anything about this definite integral? Well, what if we use the lower rectangle? This lower rectangle is going to have area because it's width is 3 and its height is 2. So the area of the lower rectangle is 6, which means that the area of the region is greater than or equal to 6. And so we know that the definite integral from 5 to 8 of g of x must be greater than or equal to 6. But since these two integrals add to 16, then if this one is greater than or equal to 6, then the other one must be less than or equal to 10. In the previous problem, we used information about the function values to find a bound for the integral. Can we go the other way and use the information about the integral to find a bound for the function values? Well, let's try it out. So let's do this problem in the hardest way possible and not draw a picture. Or maybe not. There are no bands of rogue mathematicians around me, so I'll take things the easy way and draw a picture based on what we know. So f of x is decreasing, and so we know the graph of y equals f of x is falling. The first definite integral tells us that the area under the graph from x equals 0 to x equals 5 is 15, and the second tells us that the area of the graph from 0 to 8 is 20. And that means this last portion of area from x equals 5 to x equals 8 must be 5. And so we know the value of this definite integral is also 5. So the upper rectangle for this region has width of 3 and height of f of 5. And since this must exceed the actual area, this means that 3 times f of 5 must be greater than or equal to 5. Which tells us that f of 5 is greater than or equal to 5 thirds, which means that 5 thirds is a lower bound for f of 5. That works so well. Let's see what else we can do. We know that the definite integral from 0 to 5 has value 15, and so that tells us the area from x equals 0 to x equals 5 has area 15. Well, let's take a look at the lower rectangle for this region. It's going to have width of 5 and height of f of 5. And since this must be less than the actual area, this tells us f of 5 must be less than or equal to 3. And so 3 is an upper bound for the value.