Norm Prokup

Cornell University
PhD. in Mathematics

Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.

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The Definite Integral - Problem 2

Norm Prokup
Norm Prokup

Cornell University
PhD. in Mathematics

Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.

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When a graph is a curve, find the definite integral of the function to find the area under the curve. So, antidifferentiate the function. The definite integral from points a to b is the antiderivative at b minus the antiderivative at a. For example, if looking at the function is f(x)=x2 from x=1 to x=4, the antiderivative of f(x) is x3/3. The area under the curve is 43/3 - 13/3 = 64/3 - 1/3 = 63/3 = 21.

Let's do another definite integral problem. This one is going to require a fact from Geometry. What's the area under a parabolic arch?

So imagine I construct a parabolic arch which has a shape of a parabola, and a flat base. It's kind of like a triangle except that the sides are curved. If this were a triangle, the area would be 1/2 base times height. Archimedes found that for a parabolic arch, the area is 2/3 of the base times height. So we're going to use this in the next problem.

Before I take you to the next problem, we need to review one fact about definite integrals. Now I've made the point of saying that, when the function that we're integrating is greater than or equal to 0, the definite integral gives me the exact area under the curve. However, when the function is less than or equal to 0, when the graph is below the x axis, the definite integral doesn't give me the area. It gives me the opposite of the area. Now you know that area is always positive. So this means that definite integrals are going to have negative values when you're integrating a curve that's below the x axis. So it's something to remember.

So here is the problem. The graph of y equals the quantity x minus 3 squared is shown below, this graph here. Evaluate first, the integral from 0 to 6 of that function. Well, let's do that.

First let's observe what area that represents. From 0 to 6, the interval's from here to here. Since its function is non-negative on this interval, the integral is just going to equal the area under the curve. So all we have to do is figure out what this area is. Of course that's not straightforward, because this is not a parabolic arch. However, this is a parabolic arch in here. So what we can do is find the area of this rectangle, and subtract the area of the parabolic arch. So let me write that down; equals area of rectangle minus area of I'll write para; parabolic arch.

Now the area of the rectangle is going to be the base times height. The base is 6, the height (this is not to the same scale), but the height is going to be 9. So 6 times 9 minus the area of the parabolic arch. Remember that formula; 2/3 base times height. So 2/3 (and this is the base, the flat part) the base has a length of 6. So 2/3, 6 times, (and the height is actually the same as the height of the rectangle was) it's going to be 9, the y value here. So 6 times 9. So this is 54 minus 2/3 of 54, is going to be 1/3 of 54, which is 18. This will be 54 minus 36, so that's 18. Very easy if you use subtractions. You can use the fact that the area under a parabolic arch is 2/3 base height.

Let's take a look at this slightly harder problem. The integral from 0 to 6 of this function. It's our original function minus 9. So imagine taking this red curve, and subtracting 9. If we subtract 9, these two points are going to be down here. This point will be 9 units lower, so let me translate it downwards. These two points go here. This point goes nine units down, and the graph will look something like this. So this would be the graph of this function. Let's call it g(x). So this is y equals g(x).

Now the integral from 0 to 6 of g(x), where would that be? Well, because this function lies entirely below the x axis, the value of the definite integral is going to e the opposite of the area. This is the fact that we had before. So if I calculate the area, and put a minus in front of it, that's the answer.

Well, what would this area be? Equals 18 area. What would this area be? Well, it's the same parabolic arch we had before. It's going to be 2/3 the base of 6 times the height of 9. I put a minus in front of that. So minus 2/3 base of 6 height of 9. This is -36

So the lesson to learn from this is, when you're integrating a function that's below the x axis, expect to get an answer that's negative. When you integrate a function that's entirely above the x axis, expect to get an answer that's positive.

In the first case, the integral exactly equals the area under the curve; the area between the curve, and the x axis. In the second case, the integral equals the opposite. Put a minus in front of the area between the curve, and the x axis.

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