Question

The graphs of \(y=x^{4}-2 x^{2}+1\) and \(y=1-x^{2}\) intersect at three points. However, the area between the curves can be found by a single integral. Explain why this is so, and write an integral for this area.

Short answer

The area between the graphs of the functions \(y=x^{4}-2 x^{2}+1\) and \(y=1-x^{2}\) can be calculated by the single integral \(A = \int_{-1}^{1} [(x^{4}-2 x^{2}+1) - (1-x^{2})] dx\) because the function \(y=x^{4}-2 x^{2}+1\) is always greater than or equal to the function \(y=1-x^{2}\) in the interval \([-1,1]\).

Step-by-step solution

Equating the functions

To find the points of intersection of the two curves, set the two functions \(y=x^{4}-2 x^{2}+1\) and \(y=1-x^{2}\) equal to each other. This gives us \(x^{4}-2 x^{2}+1 = 1-x^{2}\). Simplifying this equation will give us the points of intersection.

Solving the equation

Solve the derived equation from step 1, which is \(x^{4}-x^{2}=0\). Factor out \(x^{2}\) from the equation to get \(x^{2}(x^{2}-1)=0\), which gives the roots as \(x=0\) and \(x=\pm1\). These are the three intersection points of the two curves.

Plotting or analyzing the functions

It can be seen from the analysis or the plot that the function \(y=x^{4}-2 x^{2}+1\) is always greater than or equal to the function \(y=1-x^{2}\) in the interval \([-1,1]\). Therefore, the area between the two curves can be found by a single integral from -1 to 1 of the difference between the two functions.

Writing the integral for the area

The area \(A\) can be obtained by integrating the difference of the two functions from -1 to 1. So, the integration can be written as follows : \(A = \int_{-1}^{1} [(x^{4}-2 x^{2}+1) - (1-x^{2})] dx\).