91Ó°ÊÓ

An antiderivative of a function is another function whose derivative is equal to the given function. In this case, we want to find an antiderivative of $$f(x) = 1 - \frac{1}{x^2}$$. The antiderivative is of the form: $$F(x) = \int{\left(1 - \frac{1}{x^2}\right) dx}$$ We can break this into two separate integrals: $$F(x) = \int{1 dx - \int{\frac{1}{x^2} dx}}$$ The antiderivative of $$1$$ with respect to $$x$$ is $$x$$ and the antiderivative of $$\frac{1}{x^2}$$ with respect to $$x$$ is $$-\frac{1}{x}$$. Thus, $$F(x) = x - (-\frac{1}{x})$$ or $$F(x) = x + \frac{1}{x} + C$$ However, we can ignore the constant $$C$$ as it will cancel out when we apply the Fundamental Theorem of Calculus.

The Fundamental Theorem of Calculus states that, if $$F$$ is an antiderivative of $$f$$ for all $$x$$ on the interval $$[a, b]$$, then $$\int_{a}^{b} f(x) dx = F(b) - F(a)$$ In our case, the antiderivative $$F(x) = x + \frac{1}{x}$$. We need to evaluate $$F(2)$$ and $$F\left(\frac{1}{2}\right)$$: $$F(2) = 2 + \frac{1}{2} = \frac{5}{2}$$ $$F\left(\frac{1}{2}\right) = \frac{1}{2} + \frac{1}{\frac{1}{2}} = \frac{1}{2} + 2 = \frac{5}{2}$$ Now, we can find the integral by applying the Fundamental Theorem of Calculus: $$\int_{\frac{1}{2}}^{2}\left(1-\frac{1}{x^{2}}\right) dx = F(2) - F\left(\frac{1}{2}\right) = \frac{5}{2} - \frac{5}{2} = 0$$

To sketch the graph of the function $$f(x) = 1 - \frac{1}{x^2}$$, first identify some key properties of the graph. The function is continuous, and when $$x = 1$$, $$f(x) = 0$$. Moreover, since $$\lim_{x \to 0^+} f(x) = \lim_{x \to \infty} f(x) = 1$$, we have an horizontal asymptote at $$y = 1$$. Now, plot this function and identify the points $$x = \frac{1}{2}$$ and $$x = 2$$ in order to sketch the region we want to represent. Since we found that the integral evaluates to zero, it means that the net area between $$x = \frac{1}{2}$$ and $$x = 2$$ and the function is equal to zero. In conclusion, the value of the integral is 0, meaning that the area of the regions below and above the x-axis cancels out each other. And, the graph displays the function along with the region where the integral was calculated.