The graphs of y=x^4-2 x^2+1 and y=1-x^2 intersect at three points. However, the area between the

step 1 In this question, we have two equations. The first one is y is equals to x to the power 4 minus

The graphs of y=x^4-2 x^2+1 and y=1-x^2 intersect at three...

Key Concepts

Symmetry Exploitation in Integration

Symmetry can simplify the computation of areas. In this problem, because both functions are even (symmetric about the y-axis) or the region bounded by them is symmetric, the area on one side of the y-axis mirrors that on the other. This allows one to set up a single integral over the positive half of the domain and then account for symmetry (typically by doubling the integral), or to integrate over the entire interval if the roles of the functions (upper and lower) remain consistent.

Integral Setup for Area Calculation

Once the interval of integration is determined from the intersection points and the symmetry is recognized, the area is expressed as a single integral of the difference between the two functions. Despite the curves intersecting at three points, the integrand (the difference between the top and bottom functions) does not change its sign over the interval of integration. Thus, a single expression through the correct interval correctly captures the area between the curves.

Area Between Curves

This concept involves computing the region bounded by two graphs. The area is found by integrating the difference between the upper (or outer) function and the lower (or inner) function over the range where the curves form a closed region. In problems like this one, the proper identification of the functions’ roles (i.e., which is above the other) over the interval of integration is crucial to calculating the area correctly.

Intersection Points

Intersection points occur where the two functions have the same value and serve as the limits of integration when computing the area between curves. Even if the graphs intersect at more than two points, understanding the points of intersection helps in setting up the correct interval(s) over which the difference between the functions is integrated.

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