1. **State the problem:** Calculate the definite integral $$\int_{-3}^{2} (\sin x + x^{-2} + \cos x) \, dx$$. 2. **Recall the integral formulas:** - $$\int \sin x \, dx = -\cos x + C$$ - $$\int \cos x \, dx = \sin x + C$$ - $$\int x^{-2} \, dx = \int x^{-2} \, dx = -x^{-1} + C = -\frac{1}{x} + C$$ 3. **Write the integral as sum of integrals:** $$\int_{-3}^{2} \sin x \, dx + \int_{-3}^{2} x^{-2} \, dx + \int_{-3}^{2} \cos x \, dx$$ 4. **Evaluate each integral separately:** - $$\int_{-3}^{2} \sin x \, dx = [-\cos x]_{-3}^{2} = (-\cos 2) - (-\cos (-3)) = -\cos 2 + \cos 3$$ - $$\int_{-3}^{2} \cos x \, dx = [\sin x]_{-3}^{2} = \sin 2 - \sin (-3) = \sin 2 + \sin 3$$ 5. **Evaluate $$\int_{-3}^{2} x^{-2} \, dx$$:** - Note that $$x^{-2} = \frac{1}{x^2}$$ which is undefined at $$x=0$$, and the interval $$[-3,2]$$ includes 0, so the integral is improper. - We split the integral at 0: $$\int_{-3}^{2} x^{-2} \, dx = \int_{-3}^{0} x^{-2} \, dx + \int_{0}^{2} x^{-2} \, dx$$ - Evaluate each as improper integrals: $$\int_{-3}^{0} x^{-2} \, dx = \lim_{t \to 0^-} \int_{-3}^{t} x^{-2} \, dx = \lim_{t \to 0^-} [-\frac{1}{x}]_{-3}^{t} = \lim_{t \to 0^-} (-\frac{1}{t} + \frac{1}{-3}) = -\infty$$ - Since this limit diverges to negative infinity, the integral does not converge. 6. **Conclusion:** The integral $$\int_{-3}^{2} (\sin x + x^{-2} + \cos x) \, dx$$ is divergent due to the term $$x^{-2}$$ at $$x=0$$. **Final answer:** The integral does not converge (diverges).