How to find area bounded by two curves

Area between two curves

So far we have only found areas between the graph \(y=f(x)\) and the \(x\)-axis. In general we can find the area enclosed between two graphs \(y=f(x)\) and \(y=g(x)\). If \(f(x) > g(x)\), then the desired area is that which is below \(y=f(x)\) but above \(y=g(x)\), which is

\[ \int_a^b \big( f(x) - g(x) \big) \; dx. \]

This formula works regardless of whether the graphs are above or below the \(x\)-axis, as long as the graph of \(f(x)\) is above the graph of \(g(x)\). (Can you see why?)

In general, two graphs \(y=f(x)\) and \(y=g(x)\) may cross. Sometimes \(f(x)\) may be larger and sometimes \(g(x)\) may be larger. In order to find the area enclosed by the curves, we can find where \(f(x)\) is larger, and where \(g(x)\) is larger, and then take the appropriate integrals of \(f(x) - g(x)\) or \(g(x) - f(x)\) respectively.

Example

Find the area enclosed between the graphs \(y=x^2\) and \(y=x\), between \(x=0\) and \(x=2\).

Solution

Solving \(x^2 = x\), we see that the two graphs intersect at \((0,0)\) and \((1,1)\). In the interval \([0,1]\), we have \(x \geq x^2\), and in the interval \([1,2]\), we have \(x^2 \geq x\). Therefore the desired area is

\begin{align*} \int_0^1 ( x - x^2 ) \; dx \, + \int_1^2 ( x^2 - x ) \; dx &= \Bigg[ \frac{1}{2} x^2 - \frac{1}{3} x^3 \Bigg]_0^1 + \Bigg[ \frac{1}{3} x^3 - \frac{1}{2} x^2 \Bigg]_1^2 \\ &= \Bigg( \Big( \frac{1}{2} - \frac{1}{3} \Big) - \big( 0 \big) \Bigg) + \Bigg( \Big( \frac{8}{3} - 2 \Big) - \Big( \frac{1}{3} - \frac{1}{2} \Big) \Bigg) \\ &= \Big( \frac{1}{6} - 0 \Big) + \Bigg( \frac{2}{3} - \Big( {-} \frac{1}{6} \Big) \Bigg) \\ &= \frac{1}{6} + \frac{5}{6} = 1. \end{align*}

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Video transcript

- [Instructor] We have already covered the notion of area between a curve and the x-axis using a definite integral. We are now going to then extend this to think about the area between curves. So let's say we care about the region from x equals a to x equals b between y equals f of x and y is equal to g of x. So that would be this area right over here. So based on what you already know about definite integrals, how would you actually try to calculate this? Well one natural thing that you might say is well look, if I were to take the integral from a to b of f of x dx, that would give me the entire area below f of x and above the x-axis. And then if I were to subtract from that this area right over here, which is equal to that's the definite integral from a to b of g of x dx. Well then I would net out with the original area that I cared about. I would net out with this area right over here. And that indeed would be the case. And we know from our integration properties that we can rewrite this as the integral from a to b of, let me put some parentheses here, of f of x minus g of x, minus g of x dx. And now I'll make a claim to you, and we'll build a little bit more intuition for this as we go through this video, but over an integral from a to b where f of x is greater than g of x, like this interval right over here, this is always going to be the case, that the area between the curves is going to be the integral for the x-interval that we care about, from a to b, of f of x minus g of x. So I know what you're thinking, you're like okay well that worked when both of them were above the x-axis, but what about the case when f of x is above the x-axis and g of x is below the x-axis? So for example, let's say that we were to think about this interval right over here. Let's say this is the point c, and that's x equals c, this is x equals d right over here. So what if we wanted to calculate this area that I am shading in right over here? You might say well does this actually work? Well let's think about now what the integral, let's think about what the integral from c to d of f of x dx represents. Well that would represent this area right over here. And what would the integral from c to d of g of x dx represent? Well you might say it is this area right over here, but remember, over this interval g of x is below the x-axis. So this would give you a negative value. But if you wanted this total area, what you could do is take this blue area, which is positive, and then subtract this negative area, and so then you would get the entire positive area. Well this just amounted to, this is equivalent to the integral from c to d of f of x, of f of x minus g of x again, minus g of x. Let me make it clear, we've got parentheses there, and then we have our dx. So once again, even over this interval when one of, when f of x was above the x-axis and g of x was below the x-axis, we it still boiled down to the same thing. Well let's take another scenario. Let's take the scenario when they are both below the x-axis. Let's say that we wanted to go from x equals, well I won't use e since that is a loaded letter in mathematics, and so is f and g. Well let's just say well I'm kinda of running out of letters now. Let's say that I am gonna go from I don't know, let's just call this m, and let's call this n right over here. Well n is getting, let's put n right over here. So what I care about is this area, the area once again below f. We're assuming that we're looking at intervals where f is greater than g, so below f and greater than g. Will it still amount to this with now the endpoints being m and n? Well let's think about it a little bit. If we were to evaluate that integral from m to n of, I'll just put my dx here, of f of x minus, minus g of x, we already know from our integral properties, this is going to be equal to the integral from m to n of f of x dx minus the integral from m to n of g of x dx. Now let's think about what each of these represent. So this yellow integral right over here, that would give this the negative of this area. So that would give a negative value here. But the magnitude of it, the absolute value of it, would be this area right over there. Now what would just the integral, not even thinking about the negative sign here, what would the integral of this g of x of this blue integral give? Well that would give this the negative of this entire area. But now we're gonna take the negative of that, and so this part right over here, this entire part including this negative sign, would give us, would give us this entire area, the entire area. This would actually give a positive value because we're taking the negative of a negative. But if with the area that we care about right over here, the area that we cared about originally, we would want to subtract out this yellow area. Well this right over here, this yellow integral from, the definite integral from m to n of f of x dx, that's exactly that. That is the negative of that yellow area. So if you add the blue area, and so the negative of a negative is gonna be positive, and then this is going to be the negative of the yellow area, you would net out once again to the area that we think about. So in every case we saw, if we're talking about an interval where f of x is greater than g of x, the area between the curves is just the definite integral over that interval of f of x minus g of x dx.

How do you find the bounded area?