So far, we have learned how to find the integral of a function, or the area underneath a single curve. In this section, we will cover how to find the area between 2 functions, or curves.
Formula: Area Between Curves
Case 1: If we have two curves y = F(x) and y = G(x), and F(x) > G(x), this means that F(x) is the top curve and G(x) is the bottom curve, To find the area between the curves, we use the following formula: For case 1, the functions have to be in terms of x. That means y is on one side of the equation and all the x terms are on the other side of the equation. Case 2: If we have two curves x = F(y) and x = G(y), and F(y) > G(y), this means that F(y) is the right curve and G(y) is the left curve: To find the area between the curves, we use the following formula: For case 2, the functions have to be in terms of y. That means x is on one side of the equation and all the y terms are on the other side of the equation. 
For both cases, the upper and lower limits of integration will be found by the points of intersection. That means you have to set F(x) equal to G(x) (case 1) or F(y) equal to G(y) (case 2) and solve for the variable. In other cases, the limits of integration will be specified in the problem.
Guidelines for Finding the Area between Curves:

Example 1: Find the area of the region bounded by y = 5x^{2}and y = ½ x
Solution:
Identify the F(x) and G(x).
Let’s use a graphing calculator to graph both functions and shade the region in between the curves:
As you can see, F(x) is y = 5x^{2}(the top function) and G(x) is y = ½ x (the bottom function).
If limits of integration are not specified, find the points of intersection.
Since the limits are not specified, we have to find the points of intersection. Set F(x) and G(x) equal to each other and solve for x:
5 – x^{2}= ½ x
Subtract ½ x from both sides:
5x^{2}1/2x = 0
You can rewrite it in the form ax2+bx+c:
x^{2}1/2x+5 = 0
Using the quadratic formula,
x = 2.5 or x = 2
Our limits of integration are x = 2.5 and x = 2
Apply the formula:
This is case 1, so we will use the following formula:
The area between the two curves is 15.1875 square units.
Example 2: Find the area of the region bounded by y = sin(x) and y = cos(x) on the interval [0, 2π]
Solution:
Identify the F(x) and G(x).
Let’s graph both functions on the coordinate plane and shade the region in question:
Cos(x) is represented by the blue curve, and sin(x) is the green curve.
In this case, we see that there is no clear top and bottom function over the entire interval. However, we see that between the endpoints of the interval and points of intersection (highlighted yellow), the top and bottom functions are more apparent. In the first region, cos(x) is the top curve and sin(x) is the bottom curve. In the second region, sin(x) is the top curve and cos(x) is the bottom curve. In the third region, cos(x) is the top curve and sin(x) is the bottom curve. Since we have three regions of area, we need to do three separate integrals applying the formula.
Determine the limits of integration
In order to determine the limits, we need to first find the points of intersection. Set sin(x) equal to cos(x) and solve for x:
Sin(x) = cos(x)
Since this is not a polynomial equation, we cannot perform standard algebra. Therefore, we need to find x values using the unit circle that will make both sides equal to each other over the interval [0, 2π]. Using the unit circle, we can see that both cos(x) and sin(x) are both equal when x = π/4 and 5π/4.
The first interval of integration will be from [0, π/4], with F(x) = cos(x) and G(x) = sin(x).
The second interval of integration is [π/4, 5π/4] , with F(x) = sin(x) and G(x) = cos(x).
The third interval of integration is [ 5π/4, 2π] with F(x) = cos(x) and G(x) = sin(x).
Apply the formula:
Therefore, the area between the curves of y = sin(x) and y = cos(x) from [0, 2π] is 5.65684 square units.
Example 3: Find the area of the region bounded by y = 2, y = 2, x = y^{2} and y = x+ 1
Solution:
Identify the F(y) and G(y).
Let’s graph each of the functions:
This is case 2. The right function is F(y) = y^{2}. The left function, G(y) is y = x+1. However, we need to bring the x term on one side and the y term on the other to make it a function of y.
y = x+1
y – 1 = x
G(y) = y – 1
Determine the limits of integration
There are no points of intersection between F(y) and G(y), and we are not given a specific interval. However, we are told to find the area that is bounded by the four functions. Based on the diagram, the limits of integration will be y = 2 and y = 2.
Apply the formula:
c = 2
d = 2
F(y) = y^{2}
G(y) = y – 1
The area bounded in the region is 28/3 square units.
Example 4:
Find the area of the region bounded by x = y and x = y^{2}10
Solution:
Determine F(y) and G(y).
Let’s graph both of the functions:
As shown on the graph, F(y), or the right function, is x = y. The left function, or G(y) is x = y^{2}10.
Determine the limits of integration.
Since we are not given an interval of integration, we need to determine that by setting F(y) equal to G(y) and solving for y:
F(y) = G(y)
y = y^{2}10
y^{2}y 10= 0
Use the quadratic formula,
= 43.75468
The area bounded by the curves is 43.75468 square units.
Summary of Section
If we have two curves, the area bounded by both curves can be found using the following formulas: 
References:
https://www.wolframalpha.com/examples/mathematics/calculusandanalysis/applicationsofcalculus/areabetweencurves/
http://tutorial.math.lamar.edu/Classes/CalcI/AreaBetweenCurves.aspx
https://www.analyzemath.com/calculus/Integrals/area_between_curves.html
http://scidiv.bellevuecollege.edu/dh/Calculus_all/CC_4_7_FirstAppl.pdf
http://www.opentextbookstore.com/appcalc/Chapter36.pdf
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