You have learned how to calculate the area under a single curve through the application of definite integrals, but have you ever wondered how to calculate the area between two curves? The answer is probably not, but that's okay! The area between two curves is a more useful quantity than you might think. It can be used to determine figures such as the difference in energy consumption of two devices, the difference in the velocities of two particles and many other quantities. In this article, you will delve into the area between two curves, exploring the definition and the formula, covering many different examples as well as showing how to calculate the area between two polar curves.
Area Between Two Curves Definition
The area between two curves is defined as follows:
For two functions, \(f(x)\) and \(g(x)\), if \(f(x) \geq g(x)\) for all values of x in the interval \([a, \ b]\), then the area between these two functions is equal to the integral of \(f(x) - g(x)\);
So far, the area with respect to the \(x\)-axis has been discussed. What if you are asked to calculate the area with respect to the \(y\)-axis instead? In this case, the definition changes slightly:
For two functions, \(g(y)\) and \(h(y)\), if \(g(y) \geq f(x)\) for all values of \(y\) in the interval \([c, d]\), then the area between these functions is equal to the integral of \(g(y) -h(y)\).
Area Between Two Curves Formula
From the definition of the area between two curves, you know that area is equal to the integral of \(f(x)\) minus the integral of \(g(x)\), if \(f(x) \geq g(x)\) over the interval \([a,b]\). The formula used to calculate the area between two curves is thus as follows:
\[\begin{align} \text{Area } = & \int^b_a f(x) dx - \int^b_a g(x) \, \mathrm{d}x \\\end{align}\]
This can be simplified to give us the final area formula:
\[\text{Area } = \int^b_a \left ( f(x) - g(x) \right ) \, \mathrm{d}x\]
Figure 1 below illustrates the logic behind this formula.
Figure. 1- Calculating the area between two curves by subtracting the area under one curve from another. Here the area under \(g(x)=A_1\) is substracted from the area under \(f(x)=A\), the result is \(A_2\)
It may get confusing to remember which graph should be subtracted from which. You know that \(f(x)\) must be greater than \(g(x)\) over the entire interval and in the figure above, you can see that the graph of \(f(x)\) lies above the graph of \(g(x)\) over the entire interval. It can thus be said that area between two curves is equal to the integral of the equation of the top graph minus the bottom graph, or in mathematical form: \[ Area = \int_a^b( y_{\text{top}} - y_{\text{bottom}}) \, \mathrm{d}x \]
Area Between Two Curves Formula - y-axis
The formula used to calculate the area between two curves with respect to the \(y\)-axis is extremely similar to that used to calculate the area between two curves with respect to the \(x\)-axis. The formula is as follows:
\[\begin{align}\text{Area} = & \int^d_c g(y) \; dy - \int^d_c h(y) \, \mathrm{d}y \\= & \int^d_c (g(y) - h(y) ) \, \mathrm{d}y\end{align}\]
where \(g(y) \geq h(y) \) for all values of \(y\) in the interval \([c, d]\).
Since \(g(y)\) must be greater than \(h(y)\) over the entire interval \([c.d]\), you can also say that area between two curves with respect to the \(y\)-axis is equal to the integral of the graph on the right minus the graph on the left, or in mathematical form:
\[\text{Area} = \int_c^d \left (x_{\text{right}} - x_{\text{left}} \right) \, \mathrm{d}y\]
Something that you have to consider when integrating with respect to the \(y\)-axis is signed areas. Regions to the right of the \(y\)-axis will have a positive signed area, and regions to the left of the \(y\)-axis will have a negative signed area.
Consider the function \(x = g(y)\). The integral of this function is the signed area between the graph and the \(y\)-axis for \(y \in [c,d]\). The value of this signed area is equal to the value of the area to the right of the \(y\)-axis minus the value of the area to the left of the \(y\)-axis. The figure below illustrates the signed area of the function \(x = \frac{1}{4}y^2 -4\).
Figure. 2 - Signed area of the function \(x = \frac{1}{4}y^2 - 4\)
Remember that area to the left of the \(y\)-axis is negative, so when you are subtracting that area from the area to the right of the \(y\)-axis, you end up adding it back.
Area Between Two Curves Calculation Steps
There are a series of steps that you can follow that will make calculating the area between two curves relatively painless.
Step 1: Determine which function is on top. This can be done by sketching the functions or, in cases involving quadratic functions, completing the square. The sketches will not only help you determine which graph, but also helps you to see if there are any intercepts between the graphs that you should consider.
Step 2: Set up the integrals. You may have to manipulate the formula or split the functions into different intervals that fall within the original one, depending on the intersects and the interval over which you must calculate the intercept.
Step 3: Evaluate the integrals to get the area.
The next section will demonstrate how you can put these steps into practice.
Area Between Two Curves Examples
Find the area bound by the graphs \(f(x) = x + 5\) and \(g(x) = 1\) over the interval \([1, 5]\).
Solution:
Step 1: Determine which function is on top.
Figure. 3 - Graphs of \(f(x) = x+5\) and \(g(x) = 1\)
From Figure 3 it is clear that \(f(x)\) is the top graph.
It is helpful to shade in the region that you are calculating the area for, to help prevent confusion and possible mistakes.
Step 2: Set up the integrals. You have determined that \(f(x)\) lies above \(g(x)\), and you know the interval is \([1,5]\). Now you can begin substituting these values into the integral.
\[\begin{align}\text{Area} & = \int_{1}^{5} (f(x) - g(x)) \, \mathrm{d}x \\& = \int_{1}^{5} (x + 5 - 1) \, \mathrm{d}x \\& = \int_{1}^{5} (x + 4) \, \mathrm{d}x \\\end{align}\]
Step 3: Evaluate the integral.
\[\begin{align}\text{Area} & = \int_{1}^{5} (x + 5) \, \mathrm{d}x \\& = \left. \left (\frac{1}{2}x^2 + 5x \right) \right |_1^5 \\& = 28\end{align}\]
How would you calculate the area between two curves if no interval is given? This next example details how you go about doing this:
Calculate the area enclosed by the graphs of \(f(x) = -x^2 + 4x \) and \(g(x) = x^2\).
Solution:
Step 1: Determine which graph is on top. You must also determine the interval since one wasn't given.
Figure. 4 - Graphs of \(f(x) = -x^2 + 4x\) and \(g(x) = x^2\)
You can see from the sketch that an area is enclosed when the graph of \(f(x)\) lies above \(g(x)\). The interval must thus be the \(x\) values for which \(f(x) \geq g(x)\). To determine this interval, you must find the \(x\) values for which \(f(x) = g(x)\).
\[\begin{align}f(x) & = g(x) \\-x^2 + 4x & = x^2 \\2x^2 - 4x & = 0 \\x(x - 2) & = 0 \\\\\implies \qquad x = 0 &\text{ and } x = 2\end{align}\]
Step 2: Set up the integrals. The area enclosed by the graphs will be over the interval \([0,2]\).
\[\begin{align}\text{Area} & = \int_0^2 \left( f(x) - g(x) \right) \, \mathrm{d}x \\& = \int_0^2 \left( -x^2 + 4x - x^2 \right) \, \mathrm{d}x \\& = \int_0^2 \left( -2x^2 +4x \right) \, \mathrm{d}x \\\end{align}\]
STEP 3: Evaluate the integrals.
\[\begin{align}\text{Area} & = \int_0^2 \left( -2x^2 + 4x \right ) \, \mathrm{d}x \\& = \left. \left(-\frac{2}{3} x^3 + 2x^2 \right) \right |_0^2 \\& = \frac{8}{3}\end{align}\]
This example is another one involving two parabolas, but in this case, they do not intersect, and the interval is given.
Find the area of the region between the graphs of \(f(x) = -(x-6)^2 + 4\) and \(g(x) = (x-5)^2 + 7\) over the interval \([4,7]\).
Solution:
Step 1: Determine the top graph. Both functions are parabolas, so you can complete the square to determine which one lies above. In this example, they were given to you already in completed square form.
The graph of \(f(x)\) is a downturned parabola with its turning point at \((6,4)\). The graph of \(g(x)\) is an upturned parabola with its turning point at \((5,7)\). It is clear that \(g(x)\) is the graph that is above as its turning point lies at \(y= 7\) in comparison to \(f(x)\) whose turning point lies at \(y = 4\). Since \(g(x)\) is upturned and lies 3 units above \(f(x)\), which is downturned, you can see that the graphs do not intersect.
Figure. 5 - Graphs of \(f(x) = -(x- 6)^2 + 4\) and \(g(x) = (x-5)^2 + 7\)
Step 2: Set up the integral.
\[\begin{align}\text{Area} & = \int_4^7 \left( y_{\text{top}} - y_{\text{bottom}} \right) \, \mathrm{d}x \\& = \int_4^7 \left[ (x-5)^2 + 7 -(-(x-6)^2 + 4) \right] \, \mathrm{d}x \\& = \int_4^7 \left[ x^2 - 10x +25 + 7 - (-(x^2 -12x + 36) +4) \right] \, \mathrm{d}x \\& = \int_4^7 \left[ 2x^2 - 22x + 64 \right] \, \mathrm{d}x \\\end{align}\]
Step 3: Evaluate the integral.
\[\begin{align}\text{Area} & = \int_4^7 \left[ 2x^2 -22x + 64 \right] \, \mathrm{d}x \\& = \left. \left(\frac{2}{3}x^3 - 11x^2 + 64x \right) \right|_4^7 \\& = 15\end{align}\]
Another question could ask you to calculate the area between two curves over an interval where both curves lie above and below at some point. The following example demonstrates how you could solve such a question:
Calculate the area of the region bounded by the graphs of \(f(x) = -x^2 - 2x + 3\) and \(g(x) = x-1\) over the interval \([-4, 2]\).
Solution:
Step 1: Determine which graph lies above by sketching them as shown in Fig. 6 below.
Figure. 6 - Graph of a parabola and a line
It is clear from the sketch that both graphs lie above at some point in the given interval.
Step 2: Set up the integrals. In cases such as this one, where each graph lies both above and below, you must divide the area that you are calculating into separate regions. The total area between the two curves will then be equal to the sum of the areas of the separate regions.
You can see on the sketch that \(f(x)\) lies above \(g(x)\) over the interval \([-4, 1]\), so that will be the first region, \(R_1\). You can also see that \(g(x) \) lies above \(f(x)\) over the interval \([1, 2]\), so that will become the second region, \(R_2\).
\[\begin{align}\text{Area}_{R_1} & = \int_{-4}^1 \left( f(x) - g(x) \right) \, \mathrm{d}x \\& = \int_{-4}^1 \left( -(x+1)^2 + 4 - (x-1) \right) \, \mathrm{d}x \\& = \int_{-4}^1 \left( -x^2 - 2x + 3 - x + 1 \right) \, \mathrm{d}x \\& = \int_{-4}^1 \left( -x^2 - 3x + 4 \right) \, \mathrm{d}x\end{align}\]
and
\[\begin{align}\text{Area}_{R_2} & = \int_{1}^2 \left( g(x) - f(x) \right) \, \mathrm{d}x \\& = \int_{1}^2 \left( x- 1 - (-(x+1)^2 + 4 )) \right) \, \mathrm{d}x \\& = \int_{1}^2 \left( x -1 - (- x^2 - 2x + 3) \right) \, \mathrm{d}x \\& = \int_{1}^2 \left( x^2 + 3x - 4 \right) \, \mathrm{d}x\end{align}\]
Step 3: Evaluate the integrals.
\[\begin{align}\text{Area}_{R_1} & = \int_{-4}^1 \left( -x^2 - 3x + 4 \right) \, \mathrm{d}x \\& = \left. \left( -\frac{1}{3}x^3 -\frac{3}{2}x^2 + 4x \right) \right|_{-4}^{1} \\& = \frac{125}{6}\end{align}\]
and
\[\begin{align}\text{Area}_{R_2} & = \int_{1}^2 \left( x^2 + 3x - 4 \right) \, \mathrm{d}x \\& = \left. \left( \frac{1}{3}x^3 + \frac{3}{2}x^2 - 4x \right) \right|_1^2 \\& = \frac{17}{6}\end{align}\]
Step 4: Calculate the total area.
\[\begin{align}\text{Total Area} & = \text{Area}_{R_1} + \text{Area}_{R_2} \\& = \frac{125}{6} + \frac{17}{6} \\& = \frac{71}{3}\end{align}\]
Another example is as follows:
Compute the area enclosed by the graphs of \(f(x)\) and \(f(x)\) if \(h(x) = 3x^2 - 8x + 7\) and \(p(x) = x+ 1\).
Solution:
Step 1: Determine the top graph and the interval. Since you are being asked to calculate the area of the region enclosed by \(f(x)\) and \(g(x)\), you need to determine the intercepts of the graphs. The easiest way to do this is to sketch the graphs as shown in Fig. 7 below.
Figure. 7 - Areas between a line and a parabola
You can see from the sketch that an area is enclosed by the two graphs when \(g(x)\) lies above \(f(x)\). The interval for which this happens lies between the intercepts of \(f(x)\) and \(g(x)\). The interval is thus \([1,2]\).
Step 2: Set up the integral. Since \(g(x)\) lies above \(f(x)\), you shall subtract \(f(x)\) from \(g(x)\).
\[\begin{align}\text{Area} & = \int_1^2 ( g(x) - f(x)) \, \mathrm{d}x \\& = \int_1^2 ( x+1 - ( 3x^2 - 8x + 7)) \, \mathrm{d}x \\& = \int_1^2 (-3x^2 + 9x - 6) \, \mathrm{d}x \\\end{align}\]
Step 3: Evaluate the integral.
\[\begin{align}\text{Area} & = \int_1^2 ( -3x^2 + 9x -6) \, \mathrm{d}x \\& = \left. \left( -x^3 + \frac{9}{2}x^2 - 6x \right) \right|_1^2 \\& = 0.5\end{align}\]
Some questions could even ask you to calculate the area bounded by three functions, such as in the example below.
You are given the following three functions:
\[\begin{gather*}f(x) = \frac{4}{x^2} \\\\g(x) = 4x \\\\h(x) = \frac{1}{2} x\end{gather*}\]
Find the area of the region bounded by these graphs.
Solution:
The method for solving this question is similar to the one used in the example, where both graphs lay above and below over the interval. That is to say, this question is solved by dividing the total area into separate regions.
Step 1: First, sketch the graphs as shown in Fig. 8 below.
Figure. 8 - Graph of three curves: two lines and a hyperbola
You can see from the sketch that the area bound by the graphs extends over the interval \([0,2]\), but calculating the area has become more complicated as there are now three graphs involved.
The secret is to divide the area into separate regions. The sketch shows you that \(h(x)\) lies beneath both \(f(x)\) and \(g(x)\) over \([0,2]\). You now know that \(f(x)\) and \(g(x)\) are top graphs, and, through calculation or by looking at your sketch, you can show that they intersect at \((1, 4)\). The \(x\) value of the point where the graphs intersect is the place where you divide the total area into its separate regions, as shown in Fig.- 9 below.
Figure. 9 - The area enclosed by the two lines and the hyperbolas
Region \(R_1\) extends over the interval \([0,1]\) and is clearly bound on the top by the graph of \(f(x)\). Region \(R_2\) extends over the interval \([1,2]\) and is bound on top by the graph of \(f(x)\).
You can now calculate the area of regions \(R_1\) and \(R_2\) as you have clearly shown each region to have one top and one bottom graph.
Step 2: Set up the integrals.
\[\begin{align}\text{Area}_{R_1} & = \int_0^1 \left( g(x) - h(x) \right) \, \mathrm{d}x\\& = \int_0^1 \left( 4x - \frac{1}{2}x \right) \, \mathrm{d}x \\& = \int_0^1 \left( \frac{7}{2}x \right) \, \mathrm{d}x\end{align}\]
And
\[\begin{align}\text{Area}_{R_2} & = \int_1^2 \left( f(x) - h(x) \right) \, \mathrm{d}x \\& = \int_1^2 \left( \frac{4}{x^2} - \frac{1}{2}x \right) \, \mathrm{d}x\end{align}\]
Step 3: Evaluate the integrals.
\[\begin{align}\text{Area}_{R_1} & = \int_0^1 \left( \frac{7}{2}x \right) \, \mathrm{d}x \\& = \left. \left( \frac{7}{4} x^2 \right) \right|_0^1 \\& = \frac{7}{4} \\\end{align}\]
And
\[\begin{align}\text{Area}_{R_2} &= \int_1^2 \left( \frac{4}{x^2} - \frac{1}{2}x \right) \, \mathrm{d}x \\& = \left. \left( -\frac{4}{x} - \frac{1}{4}x^2 \right) \right|_1^2 \\& = \frac{5}{4}\end{align}\]
Step 4: Calculate the total area.\[\begin{align}\text{Total Area} &= \text{Area}_{R_1} + \text{Area}_{R_2} \\& = \frac{7}{4} + \frac{5}{4} \\& = 3\end{align}\]
You may be asked to calculate the area between two trigonometric curves. The following example demonstrates you solve questions of this nature.
Calculate the area enclosed by the graphs of \(f(x) = 4sin(x) \) and \(g(x) = cos(x) + 1\) for \(\pi \leq x \leq 2\pi\).
Solution:
Step 1: First, sketch the graphs. They intersect once over the given interval, at the point \((0,\pi\). You can see from the sketch that the graph of \(g(x)\) lies above the graph of \(f(x)\) across the entire interval.
Figure. 10 - Area enclosed by \(f(x)=\sin x\) and \(g(x)=\cos x+1\)
Step 2: Set up the integral. Since \(g(x)\) lies above \(f(x)\), you will need to subtract \(f(x)\) from \(g(x)\).
\[\begin{align}\text{Area} & = \int_{\pi}^{2\pi} (g(x) - f(x)) \, \mathrm{d}x \\& = \int_{\pi}^{2\pi} \left( \cos{x} + 1 - 4\sin{x} \right) \, \mathrm{d}x\end{align}\]
Step 3: Evaluate the integral.
\[\begin{align}\text{Area} & = \int_{\pi}^{2\pi} \left( \cos{x} + 1 - 4\sin{x} \right) \, \mathrm{d}x \\& = \left. \left( \sin{x} + x + 4\cos{x} \right) \right|_{\pi}^{2\pi} \\& = \pi + 8 \\ & = 11.14\end{align}\]
Area Between Two Polar Curves
The area of the region of a polar curve \(f(\theta)\) that is bounded by the rays \(\theta = \alpha\) and \(\theta = \beta\) is given by:
\[\frac{1}{2} \int_{\theta}^{\beta} r^{2} \, \mathrm{d}\theta = \frac{1}{2} \int_{\theta}^{\beta} f(\theta)^2 \, \mathrm{d}\theta\]
Then it follows that the formula to calculate the area between two polar curves is:
If \(f(\theta)\) is a continuous function, then the area bounded by a curve in polar form \(r = f(\theta)\) and the rays \(\theta = \alpha\) and \(\theta = \beta\) (with \(\alpha < \beta\)) is equal to
$$ \frac{1}{2} \int_{\alpha}^{\beta} \left (f_2(\theta)^2 - f_1(\theta)^2 \right) \, \mathrm{d}\theta $$
A more detailed explanation of the area under polar curves can be found in the article Area of Regions Bounded by Polar Curves.
Area Between Two Curves - Key takeaways
- The area between two curves with respect to the \(x\)-axis is given by \(\text{Area} = \int_a^b \left( f(x) - g(x) \right) \, \mathrm{d}x \), where:
- \(f(x) \geq g(x) \) over the interval \([a,b]\).
- The area between two curves with respect to the \(y\)-axis is given by \(\text{Area} = \int_c^d \left( g(y) - h(y) \right) \, \mathrm{d}x \), where:
- \(g(y) \geq h(y)\) over the interval \([c,d]\).
- Take the signed area into account when calculating the area between two curves with respect to the \(y\)-axis. The signed area to the left of the \(y\)-axis is negative, and the signed area to the right of the \(y\)-axis is positive.
- If no interval is given, then it can be determined by calculating the intercepts of the given graphs.
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