Area Of The Region Bounded By Several Curves

Consider the figure given in the thumbnail. Three continuous Curves f(x), g(x), and h(x) defined in the intervals [a, b], [a, c], and [c, b] respectively and intesect each other at ‘A’, ‘B’, and ‘C’. Now the area of the shaded region SACBA can be given by SACBA = SabBAa – SacCAa – ScbBCc.

Mathematically the area of the shaded region can be given by
Area = $\int ^{b}_{a}f\left( x\right) dx-\int ^{c}_{a}g\left( x\right) dx-\int ^{b}_{c}h\left( x\right) dx$

Example: Find the area of the region bounded by the curves y = 2x , y = 2x – x2 and the straight lines x = 0, x = 2.
Solution: The required area is
\begin{aligned}=\int ^{2}_{0}2^{x}dx-\int ^{2}_{0}\left( 2x-x^{2}\right) dx\\ \Rightarrow \left( \dfrac{2^{x}}{\ln 2}\right) _{0}^{2}-\left( x^{2}\right) _{0}^{2}+\left( \dfrac{x^{3}}{3}\right) _{0}^{2}=\dfrac{3}{\ln 2}-\dfrac{4}{3}\end{aligned}

Example: For what value(s) of ‘m’ does the area of the region bounded by the straight line y = mx and the curve y = x – x2 is equal to 9/2.
Solution: The curve y = x – x2 represents the parabola opening downward with vertex (1/2, 1/4) and intersect the x-axis at (0, 0) and (1, 0). The straight line y = mx intersect the parabola at (0, 0) and (1-m, m-m2). Here 1-m > 0 or < 0. So the area of the bounded region will be

\begin{aligned}\int ^{1-m}_{0}\left( x-x^{2}-mx\right) dx=\dfrac{\left( 1-m\right) ^{3}}{6}\\ \Rightarrow \dfrac{\left( 1-m\right) ^{3}}{6}=\dfrac{9}{2}\Rightarrow \left( 1-m\right) ^{3}=27\\ \Rightarrow 1-m=3\Rightarrow m=-2\end{aligned}

If (1-m) < 0, then

$\dfrac{-\left( 1-m\right) ^{3}}{6}=\dfrac{9}{2}\Rightarrow 1-m=-3\Rightarrow m=4$

Hence values of m are 4 and (-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 “Determining The Value of Unknown Parameters”.