The Integral is often called the * Area between a function and the x -axis* defines. But you can also use it to calculate the area between two functions, even if they are above or below the x -axis.

If f and g are two functions that are continuous on the interval [ a ; b ] and g ( x ) ≤ f ( x ) for all x in [ a ; b ], then the area enclosed by both functions is

## Area between two graphs

area between two functions

The simplest case is when you have two functions and the area you are looking for is just the area between the two intersections of the graphs (see graph on the right). It does not matter whether the area sought is completely either above or below the x -axis. Even if a part of the function would be below the x -axis, we could calculate the area in the same way.

As we can see from the graph, g ( x ) is the upper function and f ( x ) is the lower function.

Since the intersections of the function form the upper and lower limit of the integral, we also have to calculate the exact intersections. To do this, we set both functions equal. We then obtain:

Now we have all the data we need together. The area between the two functions is calculated by the following integral:

### Variant #2: Graphs intersect

Area between two functions,

which intersect

If two graphs intersect, from this point the lower function becomes the upper function and the upper function becomes the lower function. If we would not do this, the positive and negative area would add up and our area would be wrong. Therefore we have to swap the upper and lower functions or multiply the integral by -1. We can also just take the magnitude of the integral, and leave the order of f ( x ) and g ( x ) unchanged (but many teachers don’t like to see this, because you have to think less, even if it is mathematically sound).

We want to calculate the area between the functions f ( x ) and g ( x ) from a to b. We can do this in four steps:

**Find intersections**. For this we have to set f ( x ) = g ( x ). We number the interfaces from x 1 to x n by.**Determine upper and lower function**. This step can also be skipped, if you set the integrals in magnitude lines. When calculating the integrals, it can happen that an integral returns a negative value. But since the area is always positive, we have to make sure that all our partial integrals are also positive. To do this, we can either determine the upper and lower function and swap f ( x ) and g ( x ) each time, or we can simply put the individual integrals in magnitude bars, since the magnitude is always positive (or 0).**Set up partial integrals**. Now that we know at which points f ( x ) and g ( x ) intersect, we still need to set up the partial integrals and add them up. The integrals are set up according to the following pattern:**Calculate**. Finally, the individual integrals must be calculated and added together. The result is the area between the functions f ( x ) and g ( x ) from a to b . To show that this is an area, we have to add a unit to the result. For this we take the abbreviation "FE which is general for "area units stands.

#### Example

- f ( x ) = x ³-9 – x ²+24x-16 (blue) and
- g ( x ) = -0.5 – x ²+3 – x -2.5 (red)

calculate from 1 to 4.5.

- We set f ( x ) = g ( x ). The interfaces are:

x 1 = 1,

x 2 = 3,

x 3 = 4,5 - For the interval [1; 3] f ( x ) is the upper and g ( x ) the lower function. Therefore:

f ( x )> g ( x ) for all x ∈ [1; 3]. - With our integration limits and the intersections of the two functions, we can now set up the corresponding integrals:
- Finally we have to calculate the integrals:

## Area between a graph and the x-axis

The x -axis is also a function. It satisfies the function rule f ( x ) = 0. If you want to calculate the area between a function and the x -axis, you have to be careful, because below the x -axis the integral is negative. Therefore, similar to the second example, the zeros of the function must be calculated first.

Suppose we want to calculate the area of the function f ( x ) = x ³ – 4x from -2 to 2. First we set again the function equal to zero and calculate the zeros. These are x 1 = -2, x 2 = 0 and x 3 = 2. We can then use this to calculate the area of the function:

Since the function is point-symmetric and the magnitude of both integral limits is equal, we could have written the area as the product of a single integral: