As we all know that integration means calculating the area by dividing the region into a very large number of elementary strips and then adding up these elementary areas. At this stage, we are able to calculate the area bounded by a curve and a line between a given set of points. In the upcoming discussion, we will see how to find the enclosed area between two curves in calculus with two cases and examples.

## What is the Area Between Two Curves?

We know that the area is the quantity which is used to express the region occupied by the two-dimensional shapes in the planar lamina. In calculus, the evaluate the area between two curves, it is necessary to determine the difference of definite integrals of a function. The area between the two curves or function is defined as the definite integral of one function (say f(x)) minus the definite integral of other functions (say g(x)).Â Thus, it can be represented as the following:

**Area between two curves =Â âˆ« _{a}^{bÂ }[f(x)-g(x)]dx**

## How to Find the Area Between Two Curves?

**Case 1:Â **Consider two curves y=f(x) and y=g(x), where f(x) â‰¥ g(x) in [a,b]. In the given case, the point of intersection of these two curves can be given as x=a and x=b, by obtaining the given values of y from the equation of the two curves.

Our aim is to find the enclosed area between the two given curves. In order to do so, a thin vertical strip of width dx is taken between the lines x=a and x=b as shown in the figure. The height of this vertical strip is given as f(x)-g(x). So, the elementary area of this strip dA can be given as [f(x)-g(x)]dx.

Now, we know that the total area is made up of vary large number of such strips, starting from x=a to x=b.Hence, the total enclosed area A, between the curves is given by adding the area of all such strips between a and b:

\( A = \int\limits_{a}^b [f(x) – g(x)] dx\)

The enclosed area between two curves can also be calculated in the following manner,

A = (area bounded by the curve y=f(x), x-axis and the lines x=a and x=b) â€“ (area bounded by the curve y=g(x), x-axis and the lines x=a and x=b)

\( A = \int\limits_{a}^b f(x)dx- \int\limits_{a}^b g(x)dx = \int\limits_{a}^b [f(x) – g(x)] dx \)

where f(x)â‰¥ g(x), in [a,b]

**Case 2:** Consider another case, when two curves y=f(x) and y=g(x) are given, such that f(x) â‰¥ g(x) between x=a and x=c and f(x) â‰¤ g(x) between x=cÂ and x=b, as shown in the figure,

In this case to calculate the total area between the two curves, the sum of the areas of the region ACBDA and BPRQB is calculated i.e.

\( Total ~area = \int\limits_{a}^c [f(x) – g(x)] dx + \int\limits_{c}^b [g(x) – f(x)] dx\)

### Area Between Two Curves Example

Let us consider an example that will give a better understanding.

**Example:**

Find the area of the region bounded by the parabolas y=x^{2} and x=y^{2}.

**Solution:**

When the graph of both the parabolas is sketched we see that the points of intersection of the curves are (0,0) and (1,1) as shown in the figure below.

So, we need to find the area enclosed between these points which would give us the area between two curves. Also, in the given region as we can see,

y=x^{2}=g(x) and

x=y^{2}

or, y=âˆšx=f(x).

As we can see in the given region,

The area enclosed will be given as,

\( A = \int\limits_{0}^1 [f(x) – g(x)] dxÂ \)

\( = \int\limits_{0}^1 [\sqrt{x} – {x}^{2}] dxÂ \)

\( = [\frac{2}{3}{x}^{3/2} -\frac{{x}^{3}}{3}{]}_{0}^{1}\)

\( = \frac{2}{3}- \frac{1}{3}\)

\( = \frac{1}{3} sq.units\)

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