Chapter 6.4, Problem 44E

To determine

To find: The total area of the region between the given curve and the x -axis.

Answer to Problem 44E

The total area of the region between the curve and the x -axis is 8

Explanation of Solution

Given Information:

A given curve $y=x^3-4x,-2\le x\le 2$

Formula used:

To evaluate the area under the graph with axis, for a given interval $a\le x\le b$ , the following steps are to be followed.

1) Divide the given interval into sub-intervals such that its limits are the points at which the curve meets the x -axis.

2) Integrate the given curve over these limits.

3) Obtain the sum of absolute value of integrals to get the total area of the region between the curve and x -axis.

Calculation The graph of the given function $y=x^3-4x$ over the interval $-2\le x\le 2$ is as shown below.

  

Here, the blue region defines the interval over which the function is to be considered. From the graph, it can be seen that, the curve meets the x -axis at points -2, 0 and 2. Thus, the sub- intervals to be considered are [-2, 0] and [0, 2].

Considering the first interval [-2, 0], the integral of the given function is,

  $\begin{array}{c}{\displaystyle {\int }_{-2}^0\left(x^3-4x\right)}dx={\left(\frac{x^4}{4}-4\cdot \frac{x^2}{2}\right)}_{-2}^0\\ =0-0-\left(\frac{{\left(-2\right)}^4}{4}-4\cdot \frac{{\left(-2\right)}^2}{2}\right)\\ =-4+8\\ =4\end{array}$

Now, consider the interval [0, 2], integral over which is,

  $\begin{array}{c}{\displaystyle {\int }_0^2\left(x^3-4x\right)}dx={\left(\frac{x^4}{4}-4\frac{x^2}{2}\right)}_0^2\\ =\left(\frac{{\left(2\right)}^4}{4}-4\frac{{\left(2\right)}^2}{2}\right)-0\\ =4-8\\ =-4\end{array}$

Adding the absolute values of the integral, the total area of the region with the x-axis is given by,

  $\begin{array}{c}Totalarea=\left|4\right|+\left|-4\right|\\ =8\end{array}$