1. **Problem Statement:** Calculate the definite integrals: a) $$\int_1^3 6x \, dx$$ b) $$\int_{-2}^2 (3x^2 - 2) \, dx$$ 2. **Formula and Rules:** The definite integral of a function $$f(x)$$ from $$a$$ to $$b$$ is given by: $$\int_a^b f(x) \, dx = F(b) - F(a)$$ where $$F(x)$$ is the antiderivative of $$f(x)$$. 3. **Part a) Calculation:** - Function: $$f(x) = 6x$$ - Antiderivative: $$F(x) = \frac{6x^2}{2} = 3x^2$$ - Evaluate: $$\int_1^3 6x \, dx = F(3) - F(1) = 3(3)^2 - 3(1)^2 = 3 \times 9 - 3 \times 1 = 27 - 3 = 24$$ 4. **Part b) Calculation:** - Function: $$f(x) = 3x^2 - 2$$ - Antiderivative: $$F(x) = \frac{3x^3}{3} - 2x = x^3 - 2x$$ - Evaluate: $$\int_{-2}^2 (3x^2 - 2) \, dx = F(2) - F(-2) = (2^3 - 2 \times 2) - ((-2)^3 - 2 \times (-2))$$ $$= (8 - 4) - (-8 + 4) = 4 - (-4) = 4 + 4 = 8$$ 5. **Final Answers:** a) $$24$$ b) $$8$$