## Area Enclosed Between Two Intersecting Curves

Now we are going to find the area of the regions enclosed between two intersecting curves that are intersecting each other at the finite numbers of intervals. In these intervals they have behaviour like f(x)≥g(x) and g(x)≥f(x). In such conditions, we have to calculate the area for each individual region in which f(x)≥g(x) and g(x)≥f(x). In the end, we will Club all such areas in order to find the net area of the region enclosed between the functions f(x) and g(x) in the interval [a, b].

Now consider the figure given in the thumbnail. If S1, S2, S3,…. and so on are the area of the smaller regions shown in the figure, then the area of the region bounded between functions f(x) and g(x) in the interval [a, b] will be S. i.e. S = S1 + S2 + S3 + …

Mathematically, the area enclosed between the functions f(x) and g(x) in the interval [a, b] will be –

\begin{aligned}\int ^{b}_{a}\left| f\left( x\right) -g\left( x\right) \right| dx\\ \Rightarrow \int ^{c}_{a}\left( f\left( x\right) -g\left( x\right) \right) dx+\int ^{d}_{c}\left( g\left( x\right) -f\left( x\right) \right) dx\\ +\int ^{b}_{d}\left( f\left( x\right) -g\left( x\right) \right) dx\end{aligned}

Here ‘c’ and ‘d’ are the abscissa of the points of intersection of curves f(x) and g(x) such that c, d ∈ (a,b). Now with the help of a vertical strip, we will understand this concept. For this consider the figure given in the thumbnail.

When f(x)≥g(x): Let the width of the vertical strip be dx and height will be f(x)-g(x). So in this case the area of a single strip will be (f(x)-g(x)) dx. And the area for such region will be – $\int ^{c}_{a}\left( f\left( x\right) -g\left( x\right) \right) dx$

When g(x)≥f(x): Let the width of the vertical strip be dx and height will be g(x)-f(x). So in this case the area of a single strip will be (g(x)-f(x)) dx. And the area for such region will be – $\int ^{b}_{c}\left( g\left( x\right) -f\left( x\right) \right) dx$

Example: Find the area of the region enclosed between the curves y = sin x, y = cos x in the interval [0, π/2].
Solution: These two curves will intersect each other at x = π/4. If we consider their graphs then we can conclude that in the interval [0, π/4] cos x ≥ sin x. While sin x ≥ cos x in the interval [π/4, π/2].

\begin{aligned}\int ^{\pi /2}_{0}\left| \cos x-\sin x\right| dx\\ \Rightarrow \int ^{\dfrac{\pi }{4}}_{0}\left( \cos x-\sin x\right) dx+\int ^{\dfrac{\pi }{2}}_{\dfrac{\pi }{4}}\left( \sin x-\cos x\right) dx\\ \Rightarrow 2\left( \sqrt{2}-1\right) \end{aligned}

Example: Find the area of the region enclosed between the curves x = y2 and y = x – 2.
Solution: Since the point of intersection of these two curves are (1, -1) and (4, 2). Now the net required area A will be the sum of the areas A1 and and A2. i.e. A = A1 + A2 => \begin{aligned}A_{1}=2\int ^{1}_{0}\sqrt{x} dx=\dfrac{4}{3}\\ A_{2}=\int ^{4}_{1}\left( \sqrt{x}-\left( x-2\right) \right) dx=\dfrac{19}{6}\\ A=A_{1}+A_{2}=\dfrac{9}{2}\end{aligned}

Now You are highly recommended to attempt at least ten questions based on the theory that is discussed now.

So this is it from this tutorial. Hoping you people will attempt a few problems based on the topic discussed in this tutorial. In the next tutorial, we will discuss the topic “Area under curves using the concept of the horizontal strip”.