1. **Problem:** Calculate the definite integral $$\int_{-2}^1 \frac{4}{x^2} \, dx$$. 2. **Formula and rules:** The integral of $$\frac{1}{x^2}$$ is $$\int x^{-2} dx = -x^{-1} + C = -\frac{1}{x} + C$$. 3. **Apply the formula:** $$\int \frac{4}{x^2} dx = 4 \int x^{-2} dx = 4 \left(-\frac{1}{x}\right) + C = -\frac{4}{x} + C$$. 4. **Evaluate the definite integral:** $$\int_{-2}^1 \frac{4}{x^2} dx = \left[-\frac{4}{x}\right]_{-2}^1 = \left(-\frac{4}{1}\right) - \left(-\frac{4}{-2}\right) = -4 - 2 = -6$$. 5. **Interpretation:** The value of the integral is $$-6$$, which represents the net signed area under the curve $$y=\frac{4}{x^2}$$ from $$x=-2$$ to $$x=1$$. **Note:** The function $$\frac{4}{x^2}$$ has a discontinuity at $$x=0$$, which lies in the interval $$[-2,1]$$. This integral is improper and should be evaluated as the sum of two limits: $$\int_{-2}^1 \frac{4}{x^2} dx = \lim_{t \to 0^-} \int_{-2}^t \frac{4}{x^2} dx + \lim_{s \to 0^+} \int_s^1 \frac{4}{x^2} dx$$. Calculating these limits: - For $$\int_{-2}^t \frac{4}{x^2} dx$$: $$\left[-\frac{4}{x}\right]_{-2}^t = -\frac{4}{t} + 2$$, which tends to $$-\infty$$ as $$t \to 0^-$$. - For $$\int_s^1 \frac{4}{x^2} dx$$: $$\left[-\frac{4}{x}\right]_s^1 = -4 + \frac{4}{s}$$, which tends to $$+\infty$$ as $$s \to 0^+$$. Since both limits diverge, the integral does not converge in the usual sense. **Final answer:** The integral $$\int_{-2}^1 \frac{4}{x^2} dx$$ is improper and diverges.