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# Definite Integrals and Area - Problem 1

###### 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.

The area between a curve and the x-axis is simply the definite integral. Take the definite integral from x=a to x=b by finding the integral of the function, and finding the value of the integral at b, and subtracting the value of the integral at a.

Recall that when a curve is below the x-axis, the area is negative. So, at the points where the graph is below the x-axis, you must multiply the result of the definite integral between those points by -1 so you get a positive value for the area. If necessary, split the definite integral into parts, separately integrating the function over points where it is above the x-axis and below the x-axis. It is important to remember that when taking the definite integral, you normally do not multiply part of the integral by -1. This is only done when calculating the area bounded by the graph and the x-axis.

Let’s take a look at an area problem. First of all, remember that there are two types of area problems. In the first kind, this is the kind we’ll be dealing with, we have the area between a curve; y equals f(x) and the x axis, from x equals a to x equals b. This first case has two sub-cases. One case is were the function is greater than or equal to zero, it’s non negative. So its graph is above the x axis and when that happens, the definite integral gives us the area exactly.

However, when the function’s graph is below the x axis, when it’s less than or equal to zero, the definite integral gives me the opposite of the area. So if I’m looking for area, I’m going to have to make an adjustment at the end. I’m going to get a negative answer. I need to multiply by -1 to make it positive. So important to know because this problem has just that situation.

Here, the problem says find the area bounded between the x axis and the curve y equals ¼, the quantity x plus 2, times the quantity, x minus 4. This is a parabola, its graph looks something like this and I’m basically trying to find this area. I’m going to set up the definite integral. I know that the definite integral is going to give me a negative value, but I also know that the area is going to be the opposite of that value. So I’ll just set up the definite integral from -2 to 4 of this function. This curve here; ¼(x plus 2) (x minus 4)dx.

This will be negative, but I’ll fix it in the end. So in order to integrate, I need to multiply this out. Of course this is going to be x² plus 2x minus 4x minus 2x, minus 8. And then, that quadratic is going to be multiplied by a quarter. This ends up being 1/4x² minus 2x over 4 minus 1/2x minus 8/4 minus 2.

I’m going to use the fundamental theorem of calculus here. I need an antiderivative for this function. So what’s an antiderivative of 1/4x²? It's ¼ times 1/3x³. That’s 1/12x³ minus. What’s the antiderivative for 1/2x? It’s ½ times ½x² so 1/4x². An antiderivative for -2, -2x. So I evaluate this from -2 to 4.

First at 4; I have 1/12 of 4 to the 3rd, that’s 64. Maybe I’ll just write 64/12. Minus ¼ of 4², ¼ of 16 is 4, so minus 4. Minus 2 times 4, 8. So this takes care of this expression evaluated at 4. Now I subtract, I want this expression evaluated at -2. So -2³, is -8, so I have -8/12. I have -1/4 times 2², that’s 4, so minus ¼ times 4 is -1. And then minus 2 times -2 is plus 4.

Let’s see if we can simplify this a little bit. I have 64/12 minus -8/12. So 64/12 plus 8/12. 64 and 8 is 72, 72 over 12. That simplifies, we’ll take care of that in a second. Minus 12 minus, this is 3, so it’s going to be minus 3. So I have 6 minus 15, -9. So -9 is the value of the definite integral. But we know that this is the opposite of the area. So the area equals 9. That’s it.

You can use the definite integral to find the area between a curve and the x axis, even if that area is below the x axis. Just remember the definite integral will give you a negative value, you have to adjust to get the area.

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###### Norm Prokup

PhD. in Mathematics, University of Rhode Island

B.S. in Mechanical Engineering, Cornell University

He uses really creative examples for explaining tough concepts and illustrates them perfectly on the whiteboard. It's impossible to get lost during his lessons.

Thiswas EXCELLENT! I am a math teacher and have been looking for an easy/logical way to explain the lateral area of a cone to my students and this was incredibly helpful, thank you very much!”

I just learned more In 3 minutes of polygons here than I do in 3 weeks in my math class”

Hahaha, his examples are the same problems of my math HW!”

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