To evaluate the definite integral of $\int_{-1}^4 (4x^2 - 3x + 4) \, dx$, follow these steps:

Step 1: Compute the indefinite integral

The integrand is $4x^2 - 3x + 4$. The integral of each term is:

$\int 4x^2 \, dx = \frac{4x^3}{3}, \quad \int -3x \, dx = -\frac{3x^2}{2}, \quad \int 4 \, dx = 4x.$

Thus, the indefinite integral is:

$\int (4x^2 - 3x + 4) \, dx = \frac{4x^3}{3} - \frac{3x^2}{2} + 4x + C.$

Step 2: Evaluate the definite integral

The definite integral is:

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

Evaluate at $x = 4$:

$\frac{4(4)^3}{3} - \frac{3(4)^2}{2} + 4(4) = \frac{4 \cdot 64}{3} - \frac{3 \cdot 16}{2} + 16 = \frac{256}{3} - 24 + 16.$

Simplify:

$\frac{256}{3} - 8 = \frac{256}{3} - \frac{24}{3} = \frac{232}{3}.$

Evaluate at $x = -1$:

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

Simplify using a common denominator of 6:

$-\frac{4}{3} = -\frac{8}{6}, \quad -\frac{3}{2} = -\frac{9}{6}, \quad -4 = -\frac{24}{6}.$

Combine:

$-\frac{8}{6} - \frac{9}{6} - \frac{24}{6} = -\frac{41}{6}.$

Step 3: Subtract the results

The definite integral is:

$\frac{232}{3} - \left(-\frac{41}{6}\right) = \frac{232}{3} + \frac{41}{6}.$

Use a common denominator of 6:

$\frac{232}{3} = \frac{464}{6}, \quad \frac{464}{6} + \frac{41}{6} = \frac{505}{6}.$

Final Answer:

$\int_{-1}^4 (4x^2 - 3x + 4) \, dx = \frac{505}{6}.$

Would you like a detailed breakdown of any step?

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1. What is the process for finding indefinite integrals?
2. How do you determine common denominators in fractional arithmetic?
3. Why is a common denominator important in subtracting definite integrals?
4. What role do boundaries $x = -1$ and $x = 4$ play in definite integrals?
5. Can integrals of polynomials always be solved directly like this?

Tip: Always double-check constants of integration for indefinite integrals when solving symbolically.