Area Between Two Curves using horizontal strip

Area Between Two Curves Using Horizontal Strip

Let x = F(y) and x = G(y) are two continuous functions such that F(y) ≥ G(y), where y ∈ [c, d]. These two curves intersect each other at y = b. Now the area bounded by these two curves from y = c to y = d will be given by the definite integral. \begin{aligned} \int ^{d}_{c}| ( F\left( y\right) -G\left( y\right) | dy\\ =\int ^{b}_{c}\left( G\left( y\right) -F\left( y\right) \right) dy+\int ^{d}_{b}\left( F\left( y\right) -G\left( y\right) \right) dy\end{aligned}

Note: In the above mathematical statement, the difference of two functions is integrated from a defined lower limit to defined upper limit. This difference is always taken as positive.

Note: When we have G(y) ≥ F(y) for y ∈ [c, b], the definite integral \int ^{b}_{c}\left( G\left( y\right) -F\left( y\right) \right) dy is evaluated.

Note: When we have F(y) ≥ G(y) for y ∈ [b, d], the definite integral \int ^{d}_{b}\left( F\left( y\right) -G\left( y\right) \right) dy is evaluated.

Note: If we are not able to observe which function will lie on the right side and which function will be on left side. In such cases, we will integrate the difference of two functions within the defined limits. If the area is negative then we will take the modulus of that area.

Example: Find the area of the region enclosed by the line y = x – 1 and the parabola y2 = x + 1.
Solution: Since these two curves intersect each other at the (0, -1) and (3, 2). The straight line y = x – 1 always lies on the right side of the parabola y2 = x + 1 for the given interval. Hence the required area bounded by these two curves will be given by

\begin{aligned}\int ^{2}_{-1}\left( \left( y+1\right) -\left( y^{2}-1\right) \right) dy=\int ^{2}_{-1}\left( -y^{2}+y+2\right) dy\\ \Rightarrow -\int ^{2}_{-1}y^{2}dy+\int ^{2}_{-1}ydy+2\int ^{2}_{-1}dy=\dfrac{9}{2}\end{aligned}

Example: Find the area of the region bounded between the curves y = tan–1x , y = cot–1x and the y-axis.
Solution: Since these two curves intersect each other at the [1, π/4]. Hence the required area will be =2\int ^{\pi /4}_{0}\left( \tan y\right) dy=2\left( \ln secy\right) _{0}^{\pi /4}=\ln 2

Now attempt at least five questions based on this topic. 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 Enclosed Between Several Functions”.

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