As we all know that integration means calculating the area by dividing the region into 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.

Consider two curves y=f(x) and y=g(x), where f(x) ³ 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) – (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) ³ g(x), in [a,b]

Consider another case, when two curves y=f(x) and y=g(x) are given, such that f(x) ³ g(x) between x=a and x=c and f(x) £ g(x) between x=c 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.

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

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=Öx=f(x).

As we can see in the given region, f(x)³g(x)

The area enclosed will be given as,

\( A = \int\limits_{0}^1 [f(x) – g(x)] dx = \int\limits_{0}^1 [\sqrt{x} – {x}^{2}] dx = \frac{2}{3}{x}^{3/2} -\frac{{x}^{3}}{3}{]}_{0}^{1} = \frac{2}{3}- \frac{1}{3} = \frac{1}{3} sq.units\)

To learn more about the area under the curve download Byju’s- The Learning App.

**Practise This Question**