To evaluate the integral

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

we proceed by integrating each term separately and then applying the limits.

Step 1: Find the Antiderivative

1. For $2x^2$, the antiderivative is $\frac{2x^3}{3}$.
2. For $-3x$, the antiderivative is $-\frac{3x^2}{2}$.
3. For $4$, the antiderivative is $4x$.

Thus, the antiderivative of $2x^2 - 3x + 4$ is:

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

Step 2: Apply the Limits

Now, we evaluate this expression from $x = -1$ to $x = 3$:

$\left[\frac{2x^3}{3} - \frac{3x^2}{2} + 4x\right]_{-1}^{3}.$

At $x = 3$:

$\frac{2(3)^3}{3} - \frac{3(3)^2}{2} + 4(3) = \frac{2 \cdot 27}{3} - \frac{3 \cdot 9}{2} + 12 = 18 - 13.5 + 12 = 16.5.$

At $x = -1$:

$\frac{2(-1)^3}{3} - \frac{3(-1)^2}{2} + 4(-1) = \frac{2 \cdot (-1)}{3} - \frac{3 \cdot 1}{2} - 4 = -\frac{2}{3} - \frac{3}{2} - 4 = -\frac{29}{6}.$

Step 3: Compute the Difference

Now, subtract the value at $x = -1$ from the value at $x = 3$:

$16.5 - \left(-\frac{29}{6}\right) = 16.5 + \frac{29}{6} = \frac{99 + 29}{6} = \frac{128}{6} = \frac{64}{3}.$

Final Answer

$\int_{-1}^{3} (2x^2 - 3x + 4) \, dx = \frac{64}{3}.$

Would you like further details, or do you have any questions?

Here are some related questions to extend your understanding:

1. How do you find the antiderivative of a polynomial with more terms?
2. How can definite integrals be applied to compute the area under a curve?
3. What methods can be used if the integrand was a trigonometric function instead?
4. How would you set up an integral for a region bounded by two curves?
5. How does the Fundamental Theorem of Calculus connect differentiation and integration?

Tip: When evaluating definite integrals, carefully simplify each step, especially when substituting values to avoid errors.