Question:

The area enclosed by the curves $y = x^2$ and $y = \frac{2}{1+x^2}$ is determined by finding their intersection points and integrating the difference of the functions over the interval defined by these points. First, find the intersection points by setting the equations equal to each other: $$x^2 = \frac{2}{1+x^2}$$ $$x^2(1+x^2) = 2$$ $$x^4 + x^2 - 2 = 0$$ This is a quadratic equation in terms of $x^2$. Let $u = x^2$: $$u^2 + u - 2 = 0$$ $$(u-1)(u+2) = 0$$ So, $u = 1$ or $u = -2$. Since $u = x^2$, we have $x^2 = 1$ or $x^2 = -2$. The real solutions are given by $x^2 = 1$, which implies $x = \pm 1$. Thus, the curves intersect at $x = -1$ and $x = 1$. Next, determine which function is greater on the interval $[-1, 1]$. We can test a point within the interval, for example, $x=0$: For $y = x^2$, $y(0) = 0^2 = 0$. For $y = \frac{2}{1+x^2}$, $y(0) = \frac{2}{1+0^2} = 2$. Since $2 > 0$, the curve $y = \frac{2}{1+x^2}$ is above $y = x^2$ on the interval $[-1, 1]$. The area $A$ enclosed by the curves is given by the definite integral of the upper function minus the lower function from $x=-1$ to $x=1$: $$A = \int_{-1}^1 \left(\frac{2}{1+x^2} - x^2\right) dx$$ Since the integrand $\left(\frac{2}{1+x^2} - x^2\right)$ is an even function, we can simplify the integral using symmetry: $$A = 2 \int_0^1 \left(\frac{2}{1+x^2} - x^2\right) dx$$ Now, evaluate the integral: $$A = 2 \left[2\arctan x - \frac{x^3}{3}\right]_0^1$$ $$A = 2 \left[\left(2\arctan(1) - \frac{1^3}{3}\right) - \left(2\arctan(0) - \frac{0^3}{3}\right)\right]$$ $$A = 2 \left[\left(2\left(\frac{\pi}{4}\right) - \frac{1}{3}\right) - (0 - 0)\right]$$ $$A = 2 \left[\frac{\pi}{2} - \frac{1}{3}\right]$$ $$A = \pi - \frac{2}{3}$$