Objectives
After this section, students will be able to:
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find the area between curves by integrating along the \(x\)-axis.
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find the area between curves by integrating along the \(y\)-axis.
Section 6.1 Area Between Two Curves
In this section, we will learn how to find the area between curves by integrating along the \(x\)-axis and the \(y\)-axis.
Investigation 6.1.1.
Remember that we can find the area under the curve using the definite integral!
(a)
Solution.
Observe that \(f(x) = x^2 + 1\) is positive on the interval \([-1,3]\text{.}\) Hence, the definite integral represents the actual geometric area under the curve.
\begin{align*}
\text{Area} \amp= \int_{-1}^3 \left(x^2 + 1\right)\, dx \\
\amp= \left(\frac{x^3}{3} + x\right)\bigg|_{-1}^3 \\
\amp= \left(\frac{3^3}{3} + 3\right) - \left(\frac{(-1)^3}{3} + (-1)\right) \\
\amp= 12 - \left(-\frac{4}{3}\right) \\
\amp= \frac{40}{3}
\end{align*}
Geometrically speaking, the \(\frac{40}{3}\) is the area of the region highlighted in orange.
(b)
Solution.
Observe that \(f(x) = 2x + 4\) is positive on the interval \([-1,3]\text{.}\) Hence, the definite integral represents the actual geometric area under the curve.
\begin{align*}
\text{Area} \amp= \int_{-1}^3 \left(2x + 4\right)\, dx \\
\amp= \left(x^2 + 4x\right)\bigg|_{-1}^3 \\
\amp= \left((3)^2 + 4(3)\right) - \left (-1)^2 + 4(-1)\right) \\
\amp= 21 - \left(-3\right) \\
\amp= 24
\end{align*}
Geometrically speaking, the \(24\) is the area of the region highlighted in green.
(c)
Hint.
It is important to know what region are we finding the area. Richard will shade the region between these two curves for you in the diagram below:
Now, how should we find the area of the highlighted region?
Don’t over-complicate it! The result you obtain from part (a) and part (b) should come in useful.
Solution.
By observation, we can see that the two curves intersect at the points where \(x = -1\) and \(x = 3\text{.}\) Then we can find the area of the region highlighted in yellow by subtracting the region highlighted in orange (part a) from the region highlighted in green (part b). That is,
\begin{align*}
\text{Enclosed Area} \amp= \text{Area in green} - \text{Area in orange} \\
\amp= 24 - \frac{40}{3} \\
\amp= \frac{32}{3}
\end{align*}
The goal of this section is to find the area between curves. We are actually discussing the actual, geometric area between curves in this section. In other words, we want to figure out how much space is in between the curves. Then your final answer should always be positive (or at least zero). Area, in the geometric sense, cannot be negative.
Subsection Area between Two Curves
Based on the investigation, to find the area between curves, we need to first find the points of intersection to pin down the region. Then we will subtract the smaller area (area under the lower curve) from the larger area (area under the upper curve).
That is, we will need to observe which curve is higher and which curve is lower in order to determine the area between curves. It is always helpful to sketch the graphs first.
Example 6.1.5.
Find the area between the curves \(y = x^3 - 2x^2 + 10\) and \(y = 3x^2 + 4x - 10\text{.}\) The graph of the curves is shown below, where the red curve is \(y = x^3 - 2x^2 + 10\) and the blue curve is \(y = 3x^2 + 4x - 10\text{.}\)
Hint.
Solution.
Let’s think about the points of intersection first. It turns out that we don’t really need to know the \(y\)-values of the points of intersection since we are using \(x\) as our variable in the integral. Reading from the graph, we know that these two curves intersect at the points where \(x = -2\) and \(x = 2\text{.}\)
Next, let’s determine the upper curve and the lower curve within the interval \([-2,2]\text{.}\) Obviously, the red curve, which is the cubic function, is the upper curve, and the blue curve which is the quadratic function, is the lower curve.
If this fact doesn’t come clear, Richard always like to draw a representative rectangle in the region to see which curve stays on top and which curve is on the bottom (like what the diagram below shows).
Observe that the red curve is located on the top of this rectangle and the blue curve is located on the bottom of the rectangle. This implies that the teal rectangle has the height of
\begin{equation*}
\left(x^3 - 2x^2 + 10\right) - \left(3x^2 + 4x - 10\right)
\end{equation*}
We also know that the width of the rectangle is of \(dx\text{.}\)
Putting all these pieces together, the area of the green region is
\begin{align*}
\text{Area} \amp= \int_{-2}^2 \left(x^3 - 2x^2 + 10\right) - \left(3x^2 + 4x - 10\right)\, dx \\
\amp= \int_{-2}^2 \left(x^3 - 5x^2 - 4x + 20\right)\, dx \\
\amp= \left(\frac{x^4}{4} - \frac{5x^3}{3} - 2x^2 + 20x \right)\bigg|_{-2}^2 \\
\amp= \frac{160}{3} \\
\amp\approx 53.33 \qquad \text{unit}^2
\end{align*}
Let’s call the upper curve the \(y_\text{top}\) and the lower curve the \(y_\text{bot}\text{.}\) Then we can find the area between curves by integrating the \(y_\text{top} - y_\text{bot}\text{.}\) Let’s summarize this into a fancy theorem!
Theorem 6.1.8. Area between two curves.
Let \(y_\text{top}\) be the top curve and \(y_\text{bot}\) be the bottom curve. Then the area between the graphs is
\begin{equation*}
\text{Area} = \int_a^b \left(y_\text{top} - y_\text{bot}\right)\, dx
\end{equation*}
where \(x=a\) and \(x=b\) are the \(x\)-coordinates of the points of intersection (or the interval \([a,b]\) is given).
Things are going to be a bit messy if the curves intersect more than twice. But the idea of \(y_\text{top} - y_\text{bot}\) will remain the same.
Then our challenge here will become: which one is \(\boldsymbol{y_\text{top}}\) and which one is \(\boldsymbol{y_\text{bot}}\text{?}\)
Example 6.1.9.
Find the area between \(y=\sin(x)\) and \(y=\sin(2x)\) for the interval \([0,\pi]\text{.}\) The graph of the curves is shown below, where the blue curve is \(y=\sin(x)\) and the red curve is \(y = \sin(2x)\text{.}\)
Hint.
Solution.
Let’s follow Richard’s hint to split up the region into two pieces! Within the interval \([0,\pi]\text{,}\) the two curves intersect again. We can determine the \(x\)-value of this middle point of intersection by equating the two function outputs and solve for \(x\text{.}\)
\begin{align*}
\sin(x) \amp= \sin(2x) \qquad \qquad \\
\implies \qquad \qquad x \amp= \frac{\pi}{3}
\end{align*}
Now let’s split up the region at \(x = \frac{\pi}{3}\) and focus on one piece at a time!
In the region where \(x\) is in the interval of \([0,\frac{\pi}{3}]\text{,}\) we can determine the \(y_\text{top}\) and the \(y_\text{bot}\) by drawing a representative rectangle.
Observe that in this rectangle, the \(y_\text{top}\) is the red one and the \(y_\text{bot}\) is the blue one. Hence, the area between the two curves on \([0,\frac{\pi}{3}]\) is
\begin{align*}
\text{Area} \amp= \int_0^{\frac{\pi}{3}} \left(\sin(2x) - \sin(x)\right)\, dx \\
\amp= \left(-\frac{1}{2}\cos(2x) + \cos(x)\right)\bigg|_0^{\frac{\pi}{3}} \\
\amp= \left(-\frac{1}{2}\cos\left(\frac{2\pi}{3}\right) + \cos\left(\frac{\pi}{3}\right)\right) - \left(-\frac{1}{2}\cos(0) + \cos(0)\right) \\
\amp= \frac{1}{4} + \frac{1}{2} + \frac{1}{2} - 1 \\
\amp= 0.25
\end{align*}
Now let’s focus on the region where \(x\) is in \([\frac{\pi}{3}, \pi]\text{.}\) We can do the same trick to determine the \(y_\text{top}\) and the \(y_\text{bot}\) by drawing a representative rectangle.
Similarly, we can observe, using this rectangle, that \(y_\text{top}\) is the blue one and the \(y_\text{bot}\) is the red one. Hence, the area between the two curves on \([\frac{\pi}{3}, \pi]\) is
\begin{align*}
\text{Area} \amp= \int_\frac{\pi}{3}^\pi \left(\sin(x) - \sin(2x)\right)\, dx \\
\amp= \left(-\cos(x) + \frac{1}{2}\cos(2x)\right)\bigg|_\frac{\pi}{3}^\pi \\
\amp= \left(-\cos(\pi) + \cos(2\pi) \right) - \left(-\cos\left(\frac{\pi}{3}\right) + \frac{1}{2}\cos\left(\frac{2\pi}{3}\right)\right) \\
\amp= 1 + 1 + \frac{1}{2} - \frac{1}{4} \\
\amp= 2.25
\end{align*}
Now let’s put everything together! The area of the shaded region on the graph is
\begin{align*}
\text{Area} \amp= \int_0^{\frac{\pi}{3}} \left(\sin(2x) - \sin(x)\right)\, dx + \int_\frac{\pi}{3}^\pi \left(\sin(x) - \sin(2x)\right)\, dx \\
\amp= 0.25 + 2.25 \\
\amp= 2.5 \qquad \text{unit}^2
\end{align*}
Note 6.1.13. Is there a faster way to find the area between curves?
Observe that, in the previous example, we essentially only integrated \(\sin(x)\) and \(\sin(2x)\text{,}\) and subtract things in different order. Is there a faster way to do this?
There is actually a "checking" method and a "technology" method to help us streamline this problem.
Let’s discuss the "checking" method first. Recall we don’t want negative values in our final answer since the area we are discussing in this section is in the geometric sense. If you are having trouble deciding on the \(y_\text{top}\) and \(y_\text{bot}\text{,}\) then just pick one and observe if the final answer is positive or negative. If your answer is negative, 95% of the time, you mixed up the \(y_\text{top}\) and \(y_\text{bot}\text{.}\) If this is the case, you can just swap the \(y_\text{top}\) and \(y_\text{bot}\) in the setup and erase the negative sign in your final answer.
There is also a faster way using technology. Essentially, what we tried to avoid here is to have a negative area. Algebraically speaking, we can avoid any negative area by putting an absolute value in the integrand (this is similar to the total distance traveled idea in Section 5.6). If you want to set up the integral in one step, then you can find the area by evaluating the integral
\begin{equation*}
\int_0^\pi \left|\sin(x) - \sin(2x)\right|\, dx
\end{equation*}
Using technology, this integral equals to \(2.5\text{.}\) You can definitely check your answer using technology in one step. BUT Richard will not consider this as the "work" if you solve the question like this on the exam since Richard doesn’t see any antiderivative you found here. While he doesn’t want to micro-manage you on how you solve the problem, make sure you do know how to evaluate the integral using the FTC by first finding an antiderivative of the integrand without using a calculator. This is THE calculus step.
Things can be even more complicated if the curves given doesn’t represent \(y\) as a function of \(x\) (which means the curves given doesn’t pass the vertical line test). If this is the case, then finding the \(y_\text{top}\) and \(y_\text{bot}\) is sometimes impossible, as one curve can go both jobs.
Alternatively, instead of integrating things along the \(x\)-axis (meaning your integral will end with a \(dx\)), we can look at things sideway by integrating things along the \(\boldsymbol{y}\)-axis.
Example 6.1.14.
Find the area of the enclosed region between \(x=-\sin(2y)\) (the red curve) and \(x=\cos(y)\) (the blue curve) indicated in the figure below.
Hint.
Observe that neither of the two curves passes the vertical line test, making it insanely difficult to find the \(y_\text{top}\) and \(y_\text{bot}\text{.}\)
Note: It is totally possible to find the find the \(y_\text{top}\) and \(y_\text{bot}\) by figuring out the inverse functions with restricted domain. Then you will end up with more than two functions, and you can determine the \(y_\text{top}\) and \(y_\text{bot}\) respectively in smaller pieces. But this is algebraically intensive and Richard doesn’t even want to go with this route himself...
Instead, let’s look at things sideway. Instead of constructing a vertical skinny rectangle, let’s construct a skinny rectangle horizontally, making the height of the rectangle to be \(dy\text{,}\) as shown in the diagram below.
Now can we determine the width of the rectangle?
Solution.
Let’s first figure out the width of the rectangle!
Recall the reason why we can determine the height of the rectangle by \(y_\text{top} - y_\text{bot}\) is because we know the higher the point, the larger the \(y\)-value, so by doing the subtraction this way, we don’t end up obtainning a negative height.
Similarly, we know the \(x\)-value of a point is greater if the point is located more towards the right (so the "right-er" the point, the larger the \(x\)-value). To make sure the width is not negative, we want to determine the width of the rectangle by doing \(x_\text{right} - x_\text{left}\text{.}\)
Observe that the blue curve is "right-er" than the red curve, so \(x_\text{right} = \cos(y)\) and \(x_\text{left} = -\sin(2y)\text{.}\) Then we can find the area of the enclosed region by
\begin{equation*}
\int_{\underline{\quad}}^{\underline{\quad}} \left(\cos(y) - (-\sin(2y))\right)\, dy = \int_{\underline{\quad}}^{\underline{\quad}} \left(\cos(y) + \sin(2y)\right)\, dy
\end{equation*}
The next thing to determine here is the limits of integration. Since the variable is \(y\) in the integral now, we want to determine the region of the \(y\)-values. Graphically, we observe that the \(y\) value can go as high as \(y = \frac{\pi}{4}\) in the enclosed region. So the upper limit is \(\frac{\pi}{4}\text{.}\)
To determine the lower limit, we can do the same trick by equating \(-\sin(2y)\) and \(\cos(y)\) and solve for \(y\) within the region \((-\frac{\pi}{4},0)\text{.}\)
\begin{align*}
-\sin(2y) \amp= \cos(y) \qquad\qquad \\
\implies \qquad\qquad y \amp= -\frac{\pi}{6}
\end{align*}
Hence, the lower limit is \(-\frac{\pi}{6}\text{.}\)
Now putting everything together! We can find the area of the enclosed region.
\begin{align*}
\text{Area} \amp= \int_{-\frac{\pi}{4}}^\frac{\pi}{4} \left(\cos(y) - \sin(2y)\right)\, dy \\
\amp= \left(\sin(y) + \frac{1}{2}\cos(2y)\right)\bigg|_{-\frac{\pi}{6}}^\frac{\pi}{4} \\
\amp= \left(\sin\left(\frac{\pi}{4}\right) - \frac{1}{2}\cos\left(\frac{\pi}{2}\right)\right) - \left(\sin\left(-\frac{\pi}{6}\right) - \frac{1}{2}\cos\left(-\frac{\pi}{3}\right)\right) \\
\amp= \frac{\sqrt{2}}{2} + \frac{1}{2} + \frac{1}{4} \\
\amp= \frac{2\sqrt{2} + 3}{4} \approx 1.4571 \qquad \text{unit}^2
\end{align*}
