91Ó°ÊÓ

To find the intersection points between the line and the curve, we will set the equations of the line and the curve equal to each other: \(mx = \frac{x}{x^2+1}\) Now, we will solve this equation for \(x\). Notice that if \(x=0\), we have \(mx=0\) and \(y=0\). So we will consider \(x\neq0\): \(m = \frac{1}{x^2+1}\) Now we can solve for \(x\): \(x^2+1 = \frac{1}{m}\) \(x^2 = \frac{1}{m} - 1 = \frac{1-m}{m}\) To have a real solution for \(x\), the expression inside the square root must be nonnegative: \(\frac{1-m}{m} \geq 0\) This condition gives us the values for \(m\) that allow the line and the curve to have intersection points.

Now we will analyze the behavior of the line and the curve, considering the values of \(m\) that allow intersection points: 1. If \(01\), since the curve is always below or on the tangent line, the line and the curve don't intersect, so they don't enclose a region. To enclose a region, we need \(0

We can find the area of the enclosed region by calculating the definite integral between the intersection points \(x_1\) and \(x_2\). The area of the region is the difference between the integral of the line and the integral of the curve: \(A(m) = \int_{x_1}^{x_2} (mx-\frac{x}{x^2+1})dx\) To find \(x_1\) and \(x_2\), we can notice that \(x^2=\frac{1-m}{m}\) when they intersect. Moreover, notice that because of the symmetry of \(y=\frac{x}{x^2+1}\) in the \(y\)-axis, if \(x_1\) is a solution, then \(x_2=-x_1\). So we have: \(A(m) = \int_{-x_2}^{x_2} (mx - \frac{x}{x^2+1})dx\) Since the integrand is an odd function, the definite integral from \(-x_2\) to \(x_2\) of an odd function is always zero. So: \(A(m) = 0\) This result suggests that the region enclosed has zero area, but there is indeed an enclosed region for \(0