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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Cornell University
PhD. in Mathematics
Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.
To find the area between the curves of two graphs f(x) and g(x) from x=a to x=b, first determine which function is higher on the graph. If f(x) is the higher curve, the area between f(x) and g(x) will be the definite integral from a to b of f(x) - g(x). It is important that the higher curve is subtracting the lower curve. So, plug in these formulas. f(x) - g(x) is a new function, h(x), which you can then integrate with the usual methods and techniques.
Let’s use the definite integral to find the area between two curves. Recall that, that’s the second kind of area problem. And the way we do it is, to find the area between y equals f(x) and y equals g(x) is, we have to determine which function is above which. So if f(x) is above g(x), on some interval from a to b, then the area is going to equal the integral of the top function minus the bottom function. So it’s very important that you know which function is higher than the other. So you have to determine which is the top and which is the bottom.
Let’s take that in mind, when we look at this problem. It says find the area between y equals 8 plus 2x and y equals x³ minus 3x². I’ve kind of color-coded these, so you might be able to guess which one’s f(x) and which is g(x). But let’s just take a look.
If I call this one f(x), f(x) equals 8 plus 2x, let’s see what happens at 0 and a 4. F(0) is going to be 8; f(4) is going to be 8 plus 2 times 4, 8, so 16.
Let’s see what happens to this function. Let’s call this one g(x). So g(x) is x³ minus 3x². G(0) would be 0 and g(4) is going to be 4³, which is 64, minus 3 times 4², 3 times 16, which is 48, also 16. So what’s important about x equals 4 is that, both functions have the value of 16, so they cross here. And at 0, f(0) is 8, g(0) is 0. At least at x equals 0, f(x) is the higher of the two functions.
I’ve taken the liberty of graphing these two. You can see that from the graphs, the function, this is my f(x) here, the line is on top the entire way, from 0 to 4. My cubic function starts at 0 and decreases first but then increases up to 16 at x equals 4. So they meet at x equals 4. But, from 0 to 4, this is always the top function, this is always the bottom. You need to know that to set up the integral.
The area I’m looking for, is the area not just of this region, but also this part below. It’s important to know that if I integrate top minus bottom from 0 to 4, I will get this area exactly and it doesn’t matter that part of it is below the x axis. That’s completely unimportant. It’s always the integral of the top function minus the bottom function. Let’s set that up.
Area is going to be the integral from 0 to 4, of my top function which is f(x), 8 plus 2x minus the bottom function, which is g(x), x³ minus 3x². So I need to simplify this a little bit. Integral from 0 to 4 of, I only have 1 constant term, one x term, one x² term and one x³ term. I’m going to have 8 plus 2x, minus, minus, plus 3x², minus x ³. This is the function I need to integrate from 0 to 4.
First of all, I need an antiderivative for this. 8x, the antiderivative for 2x is x² so plus x². The antiderivative for 3x² is x³ over 3 times 3. Minus, because I have a minus here, the antiderivative I need is -1/4 x to the 4th. So I need to evaluate this from 0 to 4. Now let’s note when I do plug in 0, all of these terms are going to disappear. So I really only need to consider what happens when x equals 4.
We have 8 times 4, 32, plus 4², 16, plus 4³, 64, minus ¼ of 4 to the 4th. ¼ of 4 to the 4th, is 4³ which is 64 again. These guys cancel and I just have 32 plus 46. Remember when we evaluate this whole thing at 0, we just get minus 0. This is just 48.
So the area of this region is exactly 48. We found it by integrating the top function minus the bottom from the left bound to the right bound.
Unit
The Definite Integral