Solve the integral: \(\int_{-2}^{4} (-3x - 2) dx\)

Okay, let's solve this integral step-by-step!

1. Identify the integrand and limits: The function we need to integrate is \(f(x) = -3x - 2\), and the limits of integration are from \(-2\) to \(4\).
2. Find the antiderivative: We need to find a function \(F(x)\) such that \(F'(x) = f(x)\). Using the power rule for integration, the antiderivative of \(-3x\) is \(-3 \cdot \frac{x^{1+1}}{1+1} = -\frac{3}{2}x^2\). The antiderivative of \(-2\) is \(-2x\). So, the antiderivative is \(F(x) = -\frac{3}{2}x^2 - 2x\).
3. Apply the Fundamental Theorem of Calculus: The definite integral is given by \(F(b) - F(a)\), where \(a\) is the lower limit and \(b\) is the upper limit. In this case, \(a = -2\) and \(b = 4\).
4. Calculate F(4): \(F(4) = -\frac{3}{2}(4)^2 - 2(4) = -\frac{3}{2}(16) - 8 = -3 \times 8 - 8 = -24 - 8 = -32\).
5. Calculate F(-2): \(F(-2) = -\frac{3}{2}(-2)^2 - 2(-2) = -\frac{3}{2}(4) + 4 = -3 \times 2 + 4 = -6 + 4 = -2\).
6. Subtract F(-2) from F(4): \(F(4) - F(-2) = -32 - (-2) = -32 + 2 = -30\).

Ответ: -30