Area Of The Region Bounded By Several Curves

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”.

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