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Chapter 5: Problem 61

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

Expert verified
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

01

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.
02

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.
03

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.
04

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 : \(A = \int_{-1}^{1} [(x^{4}-2 x^{2}+1) - (1-x^{2})] dx\).

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Most popular questions from this chapter

Let \(n \geq 1\) be constant, and consider the region bounded by \(f(x)=x^{n},\) the \(x\) -axis, and \(x=1\). Find the centroid of this region. As \(n \rightarrow \infty\), what does the region look like, and where is its centroid?

(a) use a graphing utility to graph the region bounded by the graphs of the equations, \((b)\) find the area of the region, and (c) use the integration capabilities of the graphing utility to verify your results. $$ y=x^{4}-2 x^{2}, \quad y=2 x^{2} $$

Use the disk method to verify that the volume of a sphere is \(\frac{4}{3} \pi r^{3}\).

A cone of height \(H\) with a base of radius \(r\) is cut by a plane parallel to and \(h\) units above the base. Find the volume of the solid (frustum of a cone) below the plane.

Match each integral with the solid whose volume it represents, and give the dimensions of each solid. (a) Right circular cylinder (b) Ellipsoid (c) Sphere (d) Right circular cone (e) Torus (i) \(\pi \int_{0}^{h}\left(\frac{r x}{h}\right)^{2} d x\) (ii) \(\pi \int_{0}^{h} r^{2} d x\) (iii) \(\pi \int_{-r}^{r}\left(\sqrt{r^{2}-x^{2}}\right)^{2} d x\) (iv) \(\pi \int_{-b}^{b}\left(a \sqrt{1-\frac{x^{2}}{b^{2}}}\right)^{2} d x\) (v) \(\pi \int_{-r}^{r}\left[\left(R+\sqrt{r^{2}-x^{2}}\right)^{2}-\left(R-\sqrt{r^{2}-x^{2}}\right)^{2}\right] d x\)
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