Area Under The Curves When Function Changes Sign

Area Under The Curves When Function Changes Sign (f(x)≥0, f(x)≤0)

In this tutorial we will learn how to find the area of the region bounded by the continuous function which changes its sign. That means there will be a finite number of intervals in which the sign of the function may be positive and in other intervals, it can be negative as well.

So in this case if the function is behaving like this. So whatever area we are getting for the segment which lies below the x-axis will be taken as positive. Further, it will be added to the area of the region which lies above the x-axis. In this manner, we get the net area of the region bounded by the function f(x) and x-axis within the given interval.

Let the function is positive for the interval [a, c], negative for the interval [c, d] and again it is positive for the interval [d, b]. Then the required area of the region bounded by the function f(x), and x-axis between ‘a’ and ‘b’ will be given as.

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

Example: Find the area of the region bounded by the curve y = sin x and x-axis in the interval [π/2, 2π].
Solution: Since the given function y = sin x is negative in the interval [ π, 2π] and positive in the interval [π/2, π]. Hence the area of the said region will be given by-

\begin{aligned}\int ^{2\pi }_{\pi /2}sin xdx=\int ^{\pi }_{\dfrac{\pi }{2}}\sin x dx+\left| \int ^{2\pi }_{\pi }\sin x dx\right | \\ \Rightarrow -\left( cos x\right) _{\pi /2}^{\pi }+\left| -\left( \cos x\right) _{\pi }^{2\pi }\right| \\ \Rightarrow 1+\left| -2\right| =3\end{aligned}

Example: Find the area of the region bounded by the curve represented by the function f(x) = x3, the x-axis and the ordinates x = -2, and x = 1.
Solution: Since the given function f(x) = x3 is negative in the interval [ -2, 0] and positive in the interval [0, 1]. Hence the area of the required region will be given by-

\begin{aligned}\int ^{1}_{-2}x^{3}dx=\left| -\int ^{0}_{-2}x^{3}dx\right| +\int ^{1}_{0}x^{3}dx\\ \Rightarrow \left| \dfrac{1}{4}\left( x^{4}\right) _{-2}^{0}\right| +\dfrac{1}{4}\left( x^{4}\right) _{0}^{1}\\ \Rightarrow 4+\dfrac{1}{4}=\dfrac{17}{4}\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 a Region Bounded by two Non-intersecting Graphs”.

Leave a Comment

Your email address will not be published. Required fields are marked *