 ### Practical examples

Often we want to calculate the area between two curves in the plane. We might, for example, want to calculate the total increase over time of one function compared to another. Or we might want to calculate the cross-sectional area of a complicated machine part. A couple of other examples are shown in the plots below.

On the left is a Lorenz curve. It's a way of keeping track of income or wealth distribution in a country. The blue area below the curve illustrates the ability of people who already have a certain amount of income (low on the left, high on the right) to earn more money. It shows that people with the least wealth have the lowest capacity to generate more income, and those with the most income, the highest capacity to generate more. In a "perfect" world, everyone would have the capacity to earn more in proportion to what they have, but in fact, in most countries, the more you have, the easier it is to accumulate more.

The difference between the line of perfect equality and the curve is a measure of inequality called the Gini coefficient. It's the area A divided by the total area under the line of perfect equality, A = A/(A + B).

On the right is a supply and demand curve from economics. Markets for goods and services tend to reach an equilibrium point where supply meets demand at a stable price, the market price. The area between the demand curve and the market price is the consumer surplus – a count of "excess" consumers for whom there aren't enough goods. The area between the supply curve and the market price is the producer surplus – the number of excess "widgets" waiting for customers, or excess inventory. ### Calculating the area between curves

Calculating the area between two curves is pretty straightforward. The three panels below illustrate the process. On the left is a straightforward integral, which yields the yellow area under a curve of some smooth (actually differentiable) function, f(x) between x = a and x = b. The center panel shows the integral of another function, call it g(x), within the same interval, yielding the blue area. The area between those two curves (dotted) is just the difference of the two integrals,

$$\int_a^b f(x) dx - \int_a^b g(x) dx = \int_a^b [f(x) - g(x)] dx$$

There are three issues you will have to contend with that will complicate things. They are

1. You will need to decide which curve is higher. That isn't often even a big deal. If you get it backward, you'll just get a negative area, and you can just change the sign.

2. Sometimes the curves will intersect. You might only want to calculate the area enclosed by the intersections of two curves, for example, or your situation could look like the one below, with one limit of integration at an intersection point. You may have to find those points of intersection, and that will call for solid algebra skills.

3. Finally, the curves may cross in the interval in which you're interested, so you'll have to do more than one integral. All of these situations will be illustrated in examples below. ### Example 1

Simple area, non-overlapping functions

Find the area between the two curves

\begin{align} f(x) &= x^2 + 3x + 1 \\ g(x) &= x^2 + 2x + 11 \end{align}

between x = 1 and x = 3.

We don't really need to sketch a graph of these functions to know that they don't intersect. They're both parabolas with the same positive leading coefficient, and the y-intercept of the first is 2 units higher than the second. So we set up the integral

$$A = \int_1^3 g(x) - f(x) \, dx$$

where

$$g(x) - f(x) = -x + 10$$

to give an easy integral:

$$A = \int_1^3 (-x + 10) \, dx$$

Now we evaluate that integral

$$= -\frac{x^2}{2} + 10x \, \bigg|_1^3$$

... and evaluate the limits to get the area:

\begin{align} &= -\frac{9}{2} + 30 - \left( - \frac{1}{2} + 10 \right) \\ &= \bf 16\: units^2 \end{align}

Just for completeness, here's a graph of the functions of this problem and the calculated area. This is the most straightforward kind of area-between-curves problem. ### Example 2

Area enclosed by two curves

Calculate the area enclosed by the functions

\begin{align} f(x) &= x^2 - 4x + 5 \\ g(x) &= -x^2 + 6x - 4 \end{align}

The graph looks like this: In order to calculate the intersection points shown, we set the functions equal to each other and solve for x, which involves completing the square. (Don't forget completing the square!) Here's the whole procedure: First set the equation equal to zero, move the constant term to the right side and divide by the coefficient of x2:

\begin{align} 2x^2 - 10x + 9 &= 0 \\ 2x^2 - 10x &= -9 \\ x^2 - 5x &= \frac{-9}{2} \end{align}

Now add the square of ½ the coefficient of x to both sides, simplify and take the square root to find x:

\begin{align} x^2 - 5x + \frac{25}{4} &= \frac{-18}{4} + \frac{25}{4} \\ \left( x - \frac{5}{2} \right)^2 &= \frac{7}{4} \\ x &= \frac{5 ± \sqrt{7}}{2} \end{align}

That number, though nicely exact, will be a pain in the butt to plug into our integral result, so I'll calculate some decimal approximations and use them in the rest of the problem. Now the integral is

$$A = \int_{1.18}^{3.82} -2x^2 + 10x - 9 \, dx$$

... and the solution is

$$-\frac{2}{3}x^3 + 5x^2 - 9x \, \bigg|_{1.18}^{3.82}$$

which reduces to the area,

$$= 1.42 - (-4.75) = \bf 6.17 \: units^2$$

Pro tip: When calculating area between curves, units are almost always needed. Get into the habit of writing "units2" when the units aren't clear.

### Example 3

When curves cross in the interval of integration

Sometimes the curves cross in the interval of interest, yet we still want to calculate the area between the curves. This problem requires us to find all of the points of intersection and perform separate integrals for each region. Here's one such region, the region between

\begin{align} f(x) &= sin(x) \\ \\ g(x) &= -\frac{2 \sqrt{2}}{3 \pi} + \frac{2 \sqrt{2}}{3} \end{align}

The graph of these curves looks like this. Now the two curves, f(x) and g(x) intersect in three places, enclosing two regions. Notice the symmetry of this problem. The areas of the right and left enclose regions are the same, so we can (and always should) use that to our advantage. There's no need to calculate both areas. Just calculate one and multiply by 2.

The integral is set up like this:

$$A = \int_{\frac{\pi}{4}}^{\pi} [f(x) - g(x)] dx + \int_{\pi}^{\frac{5\pi}{4}} [g(x) - f(x)] dx$$

The upper and lower curves are arranged so as to give positive areas, but we take advantage of the symmetry in this case to write

$$A = 2 \int_{\frac{\pi}{4}}^{\pi} [f(x) - g(x)] dx$$

Here's the whole integral:

$$A = 2\int_{\frac{\pi}{4}}^{\pi} \left[ sin(x) - \left( -\frac{2 \sqrt{2}}{3 \pi} + \frac{2 \sqrt{2}}{3} \right) \right] \, dx$$

Now solve it to get

$$2 \left[ -cos(x) + \frac{\sqrt{2}}{3 \pi} x^2 = \frac{2 \sqrt{2}}{3} x \right]_{\frac{\pi}{4}}^{\pi}$$

Evaluate the result at the limits:

$$= 2 \left[ 1 + \frac{\sqrt{2}}{3\pi} + \frac{2\sqrt{2}}{3}\pi + \frac{\sqrt{2}}{2} - \frac{\pi\sqrt{2}}{48} + \frac{\pi\sqrt{2}}{6}\right]$$

And the area is

\begin{align} A &= 2 - \frac{2\pi\sqrt{2}}{3} + \sqrt{2} + \frac{7\pi\sqrt{2}}{24} \\ \\ &= \bf 1.748 \: units^2 \end{align}

Now if this integral had not been symmetric about x = 0, we'd have to have done two integrals shown in the first integral equation. By the way, this is a great candidate for numerical integration on a calculator – lots of chances for mistakes doing it by hand!

### Practice problems

 1 Calculate the area between the curves   $f(x) = x^2 + 3$   and   $g(x) = \frac{1}{2} x + 1$   on the interval $[-1, 3]$. The curves do not intersect on this interval, so this is one of the simplest kinds of area-between-curves problems. Solution f(x) (the parabola) is the upper function and g(x) is the lower. We have the intersection points here, so the area is a pretty simple integral: \begin{align} A &=\int_{-1}^3 [f(x) - g(x)]\,dx \\[5pt] &=\int_{-1}^3 [x^2 + 2 - \frac{1}{2} x]\, dx \\[5pt] &= \frac{x^3}{3} + 2x -\frac{x^2}{4} \, \bigg|_{-1}^3 \\[5pt] &= \frac{1}{3}(27) + 6 - \frac{9}{4} + \frac{1}{3} + 2 + \frac{1}{4} \\[5pt] &= 17 - 2 + \frac{1}{3} = 15\frac{1}{3} = \bf 15.33\: units^2 \end{align} 2 Calculate the area between the curves   $f(x) = x^2$   and   $g(x) = 3x + 1$. Try this by hand and using your calculator, and make sure that the areas agree. Solution First find the points of intersection (by equating the functions and completing the square to solve for the roots of the resulting quadratic equation): $$f(x) = x^2, \: and \: \: g(x) = 3x + 1$$ \begin{align} x^2 &= 3x + 1 \\ x^2 - 3x &= -1 \\ x^2 - 3x + \left(\frac{3}{2}\right)^2 &= -1 + \frac{9}{4} \\ \left(x - \frac{3}{2} \right)^2 &= \frac{5}{4} \\ x &= \frac{3 ± \sqrt{5}}{2} = \bf 2.6, \: 0.38. \end{align} Now find the area. The linear function is the upper function in this case. \begin{align} A &= \int_{0.38}^{2.6} (3x + 1 - x^2)\,dx \\[5pt] &= \frac{3}{2}x^2 + x - \frac{1}{3}x^3 \,\bigg|_{0.38}^{2.6} \\[5pt] &= \frac{3}{2} (2.6)^2 + 2.6 - \frac{1}{3} (2.6)^3 - \frac{3}{2} (0.38)^2 - 0.38 + \frac{1}{3} (0.38)^3 \\[5pt] &= \bf 6.26 \: units^2 \end{align} 3 Calculate the area between the curves   $f(x) = sin(x)$   and   $g(x) = 2 cos(x)$. You will have to find the points of intersection first. This is another good one to practice on your calculator, but make sure to do it by hand, too. Solution 3 First find the points of intersection (by equating the functions and solving for x): $$sin(x) = 2 cos(x) \: \longrightarrow \: \frac{sin(x)}{cos(x)} = 2$$ $$\longrightarrow \: tan(x) = 2 \: \longrightarrow \: x = tan^{-1}(2)$$ $$so \: \: x = 1.107, \frac{\pi}{2} + 1.07 + \dots$$ (because tan(x) repeats every $\pi/2$ units, so $$x = 1.107, \, 2.678$$ Now calculate the area. sin(x) > 2cos(x) in this region. \begin{align} A &= \int_{1.107}^{2.678} [sin(x) - 2 cos(x)]\,dx \\[5px] &= -cos(x) - 2 sin(x)\,\bigg|_{1.107}^{2.678} \\[5pt] &= -\left[ cos(2.678) + 2 sin(2.678) - cos(1.107) - 2 sin(1.107)\right] \\[5pt] &= -[-2.222] = \bf 2.222\: units^2 \end{align} 4 Calculate the area between the curves   $f(x) = ln(x)$   and $g(x) = x - 1$   on the interval   $[1, 5]$. Confirm for yourself that these curves intersect at  $x = 1$. Solution 4 These functions intersect at x = 1: $$ln(1) = 1 - 1 = 0$$ So the area between the curves is: \begin{align} A &= \int_1^5 [x - 1 - ln(x)]\,dx \\[5pt] &= \frac{x^2}{2} - x - xln|x| + x\,\bigg|_1^5 \\[5pt] &=\frac{x^2}{2} - x ln|x|\,\bigg|_1^5 =\frac{25}{2} -5ln(5)-\frac{1}{2} + ln(1) \\[5pt] &= 12 - 5ln(5) = \bf 3.95\: units^2 \end{align} $$\left(\text{Recall that} \; \int ln(x)\,dx = x ln(x) - x + C\right)$$  xaktly.com by Dr. Jeff Cruzan is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. © 2012, Jeff Cruzan. All text and images on this website not specifically attributed to another source were created by me and I reserve all rights as to their use. Any opinions expressed on this website are entirely mine, and do not necessarily reflect the views of any of my employers. Please feel free to send any questions or comments to jeff.cruzan@verizon.net.