## Area Between Two Non Intersecting Curves

Let f(x) and g(x) are two continuous functions in the interval [a, b] such that f(x)≥g(x) for every x∈[a, b]. Then the area of the region bounded between f(x) and g(x) will be the area of the region below the function f(x) and above the function g(x) in the interval [a, b].

Note: Here the non intersecting mean is, the curve will not be intersecting each other in the interval (a,b).

Mathematically the area of the region bounded between two non intersecting curves f(x) and g(x) such that f(x)≥g(x) for every x∈[a,b] will be given by the definite integral. $\int ^{b}_{a}\left( f\left( x\right) -g\left( x\right) \right) dx$

Concept: we will understand this with the help of vertical strip. For reference consider the strip given in the above figure (in thumbnail). Since the width of the vertical strip is pretty small so it can be taken as “dx” and the height of the vertical strip will be (f(x)-g(x)). Hence the area of the single strip will be = ((f(x)-g(x))) dx. Further the desired area will be $\int ^{b}_{a}\left( f\left( x\right) -g\left( x\right) \right) dx$

Note1: Since the area of any region can not be negative. Here the the function f(x) lies above the function g(x) hence we are calculating the Definite integral. $\int ^{b}_{a}\left( f\left( x\right) -g\left( x\right) \right) dx$

Note2: If the curve sketching is difficult and we do not know which function will lie above the other function. In this case, we will calculate the definite integral of the difference between those two functions in [a, b]. If the area appears to be negative then we will take the modulus of the area.

Example: Find the area of the region bounded by the curves y = sin x and y = cos x in [0, π/4].
Solution: Since in the interval [0, π/4], y = cos x lies above y = sin x. Hence the required area can be given by-

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

Example: Find the area of the bounded region enclosed between two non intersecting functions y = 2x and y = ln x and between the two straight lines x = 1, and x = 2.
Solution: Since in the interval [1, 2], y = 2x lies above y = ln x. Hence the required area can be given by-

\begin{aligned}\int ^{2}_{1}\left( 2^{x}-\ln x\right) dx\\ \Rightarrow \dfrac{1}{\ln 2}\left( 2^{x}\right) _{1}^{2}-\left( x\ln x-x\right) _{1}^{2}\\ \Rightarrow \dfrac{2}{\ln 2}-2\ln 2+1\end{aligned}

Now You are highly recommended to attempt at least five 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 of the region between two intersecting functions”.