We need to evaluate the definite integral: $\int_{-3}^{1} (-x^3 - x^2 + 3x + 4) \, dx$

1. Problem Description

We need to evaluate the definite integral:

$\int_{-3}^{1} (-x^3 - x^2 + 3x + 4) \, dx$

2. Solution Steps

First, we find the antiderivative of the integrand:

$\int (-x^3 - x^2 + 3x + 4) \, dx = -\int x^3 \, dx - \int x^2 \, dx + 3\int x \, dx + 4\int 1 \, dx$

Recall the power rule for integration:

$\int x^n \, dx = \frac{x^{n+1}}{n+1} + C$

Applying the power rule, we get:

$-\frac{x^4}{4} - \frac{x^3}{3} + \frac{3x^2}{2} + 4x + C$

Now we evaluate the definite integral:

$\int_{-3}^{1} (-x^3 - x^2 + 3x + 4) \, dx = \left[-\frac{x^4}{4} - \frac{x^3}{3} + \frac{3x^2}{2} + 4x\right]_{-3}^{1}$

$= \left(-\frac{1^4}{4} - \frac{1^3}{3} + \frac{3(1)^2}{2} + 4(1)\right) - \left(-\frac{(-3)^4}{4} - \frac{(-3)^3}{3} + \frac{3(-3)^2}{2} + 4(-3)\right)$

$= \left(-\frac{1}{4} - \frac{1}{3} + \frac{3}{2} + 4\right) - \left(-\frac{81}{4} - \frac{-27}{3} + \frac{3(9)}{2} - 12\right)$

$= \left(-\frac{1}{4} - \frac{1}{3} + \frac{3}{2} + 4\right) - \left(-\frac{81}{4} + 9 + \frac{27}{2} - 12\right)$

$= \left(-\frac{1}{4} - \frac{1}{3} + \frac{3}{2} + 4\right) - \left(-\frac{81}{4} + \frac{27}{2} - 3\right)$

$= -\frac{1}{4} - \frac{1}{3} + \frac{3}{2} + 4 + \frac{81}{4} - \frac{27}{2} + 3$

$= \left(-\frac{1}{4} + \frac{81}{4}\right) - \frac{1}{3} + \left(\frac{3}{2} - \frac{27}{2}\right) + (4+3)$

$= \frac{80}{4} - \frac{1}{3} - \frac{24}{2} + 7$

$= 20 - \frac{1}{3} - 12 + 7$

$= 15 - \frac{1}{3}$

$= \frac{45}{3} - \frac{1}{3} = \frac{44}{3}$

3. Final Answer

$\frac{44}{3}$