91Ó°ÊÓ

In order to find the antiderivative $$F(x)$$ of $$f(x) = x^2 - 4$$, we'll need to integrate the function. To do this, we'll follow the power rule of integration, which states that: $$\int x^n dx = \frac{x^{(n+1)}}{(n+1)} + C$$ (with \(C\) being a constant) Now, we will integrate $$f(x)$$: $$\int (x^2 - 4) dx = \int x^2 dx - \int 4 dx$$ Applying the power rule to $$\int x^2 dx$$ and the constant rule to $$\int 4 dx$$, we get: $$\frac{x^{(2+1)}}{2+1} - 4x + C = \frac{x^3}{3} - 4x + C$$ So, the antiderivative $$F(x)$$ of $$f(x) = x^2 - 4$$ is: $$F(x) = \frac{x^3}{3} - 4x + C$$

Now that we have the antiderivative $$F(x)$$, we can use the Fundamental Theorem of Calculus to evaluate the integral of $$f(x)$$ on the interval $$[-2, 2]$$: $$\int_{-2}^{2} \left(x^2 - 4\right) dx = F(2) - F(-2)$$ Plugging in the values into the antiderivative expression $$F(x) = \frac{x^3}{3} - 4x + C$$, we get: $$F(2) = \frac{2^3}{3} - 4(2) + C = \frac{8}{3} - 8 + C$$ $$F(-2) = \frac{(-2)^3}{3} - 4(-2) + C = \frac{-8}{3} + 8 + C$$ Now, we can subtract $$F(-2)$$ from $$F(2)$$ to find the integral: $$\int_{-2}^{2} \left(x^2 - 4\right) dx = F(2) - F(-2) = \left(\frac{8}{3} - 8 + C\right) - \left(\frac{-8}{3} + 8 + C\right)$$ $$= \frac{8}{3} - 8 + C + \frac{8}{3} - 8 - C$$ $$= \frac{16}{3} - 16$$Since the terms involving the constant $$C$$ cancel each other out in the subtraction, the integral evaluates to: $$\int_{-2}^{2} \left(x^2 - 4\right) dx = \frac{16}{3} - 16$$