14.02 Area between curves
We have looked at using integration to find the area between a curve and an axis. Now, we will consider finding the area bounded by multiple curves. There are a number of scenarios to consider and it is useful to remember areas between and under curves can be calculated by using any combination of the following:
- Standard formulas for known geometrical shapes (rectangles, triangles, trapeziums)
- Definite integrals
- Breaking a region into components and adding or subtracting one area from another
Areas between multiple curves and an axis
Sometimes we are required to find the area formed between multiple curves and an axis. The key to solving these problems is finding any points of intersection, dividing the area into sections, then evaluating and summing the integrals associated with each section.
Worked examples
Example 1
By graphing the curves $y=2x$y=2x and $y=2\left(x-2\right)^2$y=2(x−2)2, find the area bounded by the curves and the $x$x-axis.
Think: This problem involves multiple curves and hence requires graphing to see what the area looks like.
Do: By graphing both curves and shading the region of interest we get the following graph.
We can see from the graph that there is a point of intersection at $x=1$x=1. We could also determine this by solving the two functions simultaneously.
To calculate this area we need to divide it up into two sections, each section equivalent to the area underneath a function. We will calculate the area under the function $y=2x$y=2x between $0$0 and $1$1 and add this to the area under the function $y=2\left(x-2\right)^2$y=2(x−2)2 between $1$1 and $2$2.
We can setup the following integral to calculate the area, noting that the point of intersection becomes a boundary value of both integrals:
| Area | $=$= | $\int_0^12xdx+\int_1^22\left(x-2\right)^2dx$∫102xdx+∫212(x−2)2dx |
| $=$= | $\left[x^2\right]_0^1+\left[\frac{2}{3}\left(x-2\right)^3\right]_1^2$[x2]10+[23(x−2)3]21 | |
| $=$= | $1-0+0-\left(-\frac{2}{3}\right)$1−0+0−(−23) | |
| $=$= | $1+\frac{2}{3}$1+23 | |
| $=$= | $\frac{5}{3}$53 units2 |
Hence, the area bound by the two curves and the $x$x-axis is $\frac{5}{3}$53 units2.
Example 2
Find the shaded area shown below which is bounded by the curve $y=x^2$y=x2, the $y$y-axis and the line $y=4$y=4 for $x\ge0$x≥0.
Think: We can see that we could subtract the area below the blue curve (the blue area labelled $B$B in the diagram below) from the area of the rectangle formed by the $x$x and $y$y axes and the lines $x=2$x=2 and $y=4$y=4. This could also be considered as the area under the line $y=4$y=4 from $x=0$x=0 and $x=2$x=2 minus the area under $y=x^2$y=x2 over the same interval.
Do:
| $A$A | $=$= | Area of rectangle $-\int_0^2x^2dx$−∫20x2dx |
| $=$= | $2\times4-\left[\frac{x^3}{3}\right]_0^2$2×4−[x33]20 | |
| $=$= | $8-\left(\frac{8}{3}-0\right)$8−(83−0) | |
| $=$= | $5\frac{1}{3}$513 units2 |
Thus, the area in question is $5\frac{1}{3}$513 units2.
Practice questions
Question 1
Find the area bounded between the two curves $y=\sqrt{x+5}$y=√x+5 and $y=-x-3$y=−x−3 and the $x$x-axis.
Question 2
The diagram shows the shaded region bounded by $y=3$y=3, $y=0$y=0, $y=6x-x^2-8$y=6x−x2−8, $x=0$x=0 and $x=6$x=6.
Find the area of the shaded region by appropriately subtracting one area from another.
Areas between two curves
When calculating the area between two curves $f\left(x\right)$f(x) and $g\left(x\right)$g(x) and the lines $x=a$x=a and $x=b$x=b, we subtract the integral of the bottom curve from the integral of the top curve.
For example, in the diagram above we can calculate the value of the shaded area using:
| Area | $=$= | $\int_a^bf\left(x\right)dx-\int_a^bg\left(x\right)dx$∫baf(x)dx−∫bag(x)dx |
| $=$= | $\int_a^b\left(f\left(x\right)-g\left(x\right)\right)dx$∫ba(f(x)−g(x))dx |
Notice that since the endpoints are the same, calculating and subtracting the separate integrals is the same as calculating the integral of the difference of the two functions.
It may be necessary to break up the interval into sections where a particular function is on top, such as in the following diagram.
The shaded region in the diagram above can be calculated as follows:
Area $=\int_a^bf\left(x\right)-g\left(x\right)dx+\int_b^cg\left(x\right)-f\left(x\right)dx+\int_c^df\left(x\right)-g\left(x\right)dx$=∫baf(x)−g(x)dx+∫cbg(x)−f(x)dx+∫dcf(x)−g(x)dx
Finding the area between the two functions is equivalent to finding the area between the difference function $y=f\left(x\right)-g\left(x\right)$y=f(x)−g(x) and the $x$x-axis. The difference function and associated area for the example given above is shown below.
Worked examples
EXAMPLE 3
By graphing the curves $y=6-2x$y=6−2x and $y=x\left(x-3\right)^2$y=x(x−3)2, find the area bounded by the two curves.
Think: Graphing gives us the following two regions trapped between the curves. Notice that on the left the cubic function is uppermost while on the right the linear function is uppermost. The points of intersection $\left(1,4\right)$(1,4), $\left(2,2\right)$(2,2) and $\left(3,0\right)$(3,0) can be found algebraically or using a calculator.
Do: Because we need the area under the upper curve minus the area under the lower curve at all times, we need to create two calculations to accommodate where the curves switch positions at the intersection point. Setting the integral up in two parts as follows:
| Area | $=$= | $\int_1^2\left(\left(x^3-6x^2+9x\right)-\left(6-2x\right)\right)dx+\int_2^3\left(\left(6-2x\right)-\left(x^3-6x^2+9x\right)\right)dx$∫21((x3−6x2+9x)−(6−2x))dx+∫32((6−2x)−(x3−6x2+9x))dx |
| $=$= | $\int_1^2\left(x^3-6x^2+11x-6\right)dx+\int_2^3\left(6-11x+6x^2-x^3\right)dx$∫21(x3−6x2+11x−6)dx+∫32(6−11x+6x2−x3)dx | |
| $=$= | $\left[\frac{x^4}{4}-2x^3+\frac{11x^2}{2}-6x\right]_1^2+\left[6x-\frac{11x^2}{2}+2x^3-\frac{x^4}{4}\right]_1^2$[x44−2x3+11x22−6x]21+[6x−11x22+2x3−x44]21 | |
| $=$= | $-2+\frac{9}{4}+\frac{9}{4}-2$−2+94+94−2 | |
| $=$= | $\frac{1}{4}+\frac{1}{4}$14+14 | |
| $=$= | $\frac{1}{2}$12 units2 |
Thus, the area of the shaded region is $\frac{1}{2}$12 units2.
Reflect: If we had recognised that each shaded portion was the same in shape and size (a reflection of each other, or a rotation) we could have calculated one area and then multiplied by $2$2, saving us some time.
Example 4
Consider the area bounded by the graphs $y=\cos x$y=cosx and $y=\cos2x$y=cos2x on the graph below.
Think: To find the shaded area, we need to find the area under the top curve, $y=\cos x$y=cosx, and subtract from that the area under the bottom curve, $y=\cos2x$y=cos2x.
Do: We can calculate as follows:
| Area | $=$= | $\int_0^{\frac{2\pi}{3}}\left(\cos x-\cos2x\right)dx$∫2π30(cosx−cos2x)dx |
| $=$= | $\left[\sin x-\frac{\sin2x}{2}\right]_0^{\frac{2\pi}{3}}$[sinx−sin2x2]2π30 | |
| $=$= | $\left(\frac{\sqrt{3}}{2}+\frac{\sqrt{3}}{4}\right)-\left(0-0\right)$(√32+√34)−(0−0) | |
| $=$= | $\frac{3\sqrt{3}}{4}$3√34 units2 |
Hence, the shaded area is $\frac{3\sqrt{3}}{4}$3√34 units2.
Practice questions
Question 3
Consider the functions $y=-x^2+8$y=−x2+8 and $y=-x+2$y=−x+2.
Graph the functions on the axes below.
Loading Graph...State the values of $x$x at which the line and the parabola intersect.
Write both values on the same line, separated by a comma.
Hence find the area enclosed between the line and the curve.
Question 4
The figure shows the area bound by the curves $y=4\cos x$y=4cosx, $y=-4\cos\left(x-\frac{\pi}{2}\right)$y=−4cos(x−π2), the $y$y-axis and $x=\frac{3\pi}{4}$x=3π4.
Determine the shaded area.
Question 5
A diagram of the graphs of $y=\sin x$y=sinx and $y=3\sin x$y=3sinx are shown.
What integral could be performed to calculate the shaded area bound by the curves, $A$A.
Select all that apply.
$\int_0^{\pi}\left(3\sin x-\sin x\right)dx$∫π0(3sinx−sinx)dx
A$\int_0^{\pi}3\sin xdx-\int_0^{\pi}\sin xdx$∫π03sinxdx−∫π0sinxdx
B$\int_0^{\pi}3\sin xdx+\int_0^{\pi}\sin xdx$∫π03sinxdx+∫π0sinxdx
C$2\int_0^{\pi}\sin xdx$2∫π0sinxdx
D$\int_0^{\pi}\left(3\sin x+\sin x\right)dx$∫π0(3sinx+sinx)dx
E$4\int_0^{\pi}\sin xdx$4∫π0sinxdx
FCalculate the shaded area $A$A.
What is the ratio of the areas $A:B$A:B?
Using the calculator
Just as when calculating the area between two functions by hand, over an interval where $f\left(x\right)\ge g\left(x\right)$f(x)≥g(x), we can graph the functions separately and find the area under each curve over the interval and then subtract the area under $g\left(x\right)$g(x) from the area under $f\left(x\right)$f(x). Or we can find the area under the difference function, $y=f\left(x\right)-g\left(x\right)$y=f(x)−g(x), over the given interval. That is:
| Area | $=$= | $\int_a^bf\left(x\right)dx-\int_a^bg\left(x\right)dx$∫baf(x)dx−∫bag(x)dx |
| $=$= | $\int_a^b\left(f\left(x\right)-g\left(x\right)\right)dx$∫ba(f(x)−g(x))dx |
When finding the area between curves over an interval where the functions cross over, we can choose to sketch the function and calculate the area of each region separately before summing the areas, or we can integrate the absolute value of the difference function. That is, the area between the functions $f\left(x\right)$f(x) and $g\left(x\right)$g(x) on the interval $a\le x\le b$a≤x≤b is:
Area $=\int_a^b\mid f\left(x\right)-g\left(x\right)\mid dx$=∫ba∣f(x)−g(x)∣dx