Chapter 7.2, Problem 8E

To determine

To find: The area of the region enclosed by the graph of the two functions.

Answer to Problem 8E

The area is : $A\approx 4.332$

Explanation of Solution

Given information:

The functions are:

  $\begin{array}{c}y=\cos (2x),\\ y=x^2-2\end{array}$

Calculation:

First determine the limits of integration before determine the size of the region bounded by these two functions. In other words, equate to determine the intersecting points:

  $\cos (2x)=x^2-2$

However, see, this problem cannot be solved using elementary techniques, so use the calculator's function. Consequently, to use the following code

  $\text{Intersect }\left(\cos (2x),x^2-2,-2,2\right)$

Get the values,

  $\begin{array}{c}x=-1.153\\ x=\mathrm{1.153.}\end{array}$

Hence,

  $\cos (2x)\ge x^2-2$

For each

  $x\in [-1.153,1.153]$

To use the definition and get:

  $\begin{array}{c}A={\int }^{\text{}}{}_{-1.153}^{1.153}\left[\cos (2x)-\left(x^2-2\right)\right]\text{d}x\\ ={\int }^{\text{}}{}_{-1.153}^{1.153}\cos (2x)\text{d}x-{\int }^{\text{}}{}_{-1.153}^{1.153}{x}^2\text{d}x+2{\int }^{\text{}}{}_{-1.153}^{1.153}\text{d}x\\ =\frac{1}{2}{\int }^{\text{}}{}_{-2.306}^{2.306}\cos t\text{d}t-{\left.\frac{x^3}{3}\right|}_{-1.153}^{1.153}+{\left.2x\right|}_{-1.153}^{1.153}\\ ={\left.\frac{1}{2}\sin t\right|}_{-2.306}^{2.306}-\frac{1}{3}\left({1.153}^3-{(-1.153)}^3\right)+2\cdot (1.153-(-1.153))\\ \approx \frac{1}{2}(\sin (2.306)-\sin (-2.306))-3.59\\ \approx 0.742+3.59\\ =4.332\end{array}$

For first integral to use the substitution $2x=t$ where is

  $\text{d}x=\frac{1}{2}\text{d}t$ and limits of integration are now $\pm 2.306$ .

Lets verify this result by enter the code:

Integral between $\left(\cos (2x),x^2-2,-1.153,1.153\right)$

The graph is shown below:

  

Therefore, the required area of the region enclosed by the graph of the two functions is

  $A\approx 4.332$ .