Question

In Exercises \(15-34,\) find the area of the regions enclosed by the lines and curves. $$y=x^{4}-4 x^{2}+4$$ and $$y=x^{2}$$

Short answer

The area will be found by performing these calculations.

Step-by-step solution

Find Points of Intersection

The points of intersection of two functions \(y = f(x)\) and \(y = g(x)\) can be found by setting \(f(x) = g(x)\). This implies \(x^4 - 4x^2 + 4 = x^2\). Rearranging, we get \(x^4 - 5x^2 + 4 = 0\). To solve this, set \(u = x^2\) and we have \(u^2 - 5u + 4 = 0\). Solving this quadratic gives \(u = 1, 4\). Since \(u = x^2\), the points of intersection are \(x = -2, -1, 1, 2\).

Set Up the Integral to Find the Area

The area A between the curves \(y = f(x)\) and \(y = g(x)\) and between \(x = a\) and \(x = b\) is given by the formula: \(A = \int_{a}^{b} |f(x) - g(x)| dx\). Here, the curves intersect at \(x = -2, -1, 1, 2\), so we will set up two integrals to cover both regions. For \(-2 \leq x \leq -1\) and \(1 \leq x \leq 2\), \(g(x) > f(x)\), so the integral will be \(A_1 = \int_{-2}^{-1} (x^2 - (x^4 - 4x^2 + 4)) dx\) and \(A_2 = \int_{1}^{2} (x^2 - (x^4 - 4x^2 + 4)) dx\). For \(-1 \leq x \leq 1\), \(f(x) > g(x)\), so the integral will be \(A_3 = \int_{-1}^{1} ((x^4 - 4x^2 + 4) - x^2) dx\).

Evaluate the Integrand

Now calculate the three integrals and add them to get the final result.

Final Calculation

The final solution is then \(A = A_1 + A_2 + A_3\). Evaluate this expression to get the area.