Chapter 7.2, Problem 26E

To determine

To find: The area of the region enclosed by the lines and curve.

Answer to Problem 26E

The area is: $A=6\frac{14}{15}$

Explanation of Solution

Given information:

The curves are:

  $\begin{array}{c}4x^2+y=4\\ x^4-y=1\end{array}$

Calculation:

To find the area of the region which is enclosed by the curves are:

  $\begin{array}{c}4x^2+y=\mathrm{4.....}(1)\\ x^4-y=\mathrm{1.....}(2)\end{array}$

Find the intersection points, i.e. the limits of integrations. Therefore,

  $\begin{array}{c}{x}^4-1=4-4x^2\\ \Rightarrow x^4+4x^2-5=0\\ \Rightarrow x^4-x^2+5x^2-5=0\\ \Rightarrow x^2\left(x^2-1\right)+5\left(x^2-1\right)=0\\ \Rightarrow \left(x^2+5\right)\left(x^2-1\right)=0\\ \Rightarrow x^2-1=0\\ \Rightarrow x=\pm 1\end{array}$

The interval of integration is $[-1,1]$ .

Since, $4-4x^2\ge x^4-1$

For each ,

  $x\in [-1,1]$

To use the definition and get:

  

  $\begin{array}{c}={\left.\left[5x-4\cdot \frac{x^3}{3}-\frac{x^5}{5}\right]\right|}_{-1}^1\\ =5(1+1)-\frac{4(1+1)}{3}-\frac{1+1}{5}\\ =10-\frac{8}{3}-\frac{2}{3}\\ =\frac{104}{15}\\ =6\frac{14}{15}\end{array}$

Therefore, the required area of the region enclosed by the lines and curve is $6\frac{14}{15}$ .