Let's solve each definite integral step-by-step:

1. $\int_2^3 x^2 \, dx$

To evaluate this integral, find the antiderivative of $x^2$, which is $\frac{x^3}{3}$.

$\int_2^3 x^2 \, dx = \left[ \frac{x^3}{3} \right]_2^3 = \frac{3^3}{3} - \frac{2^3}{3} = \frac{27}{3} - \frac{8}{3} = \frac{19}{3}$

2. $\int_4^5 e^x \, dx$

The antiderivative of $e^x$ is $e^x$.

$\int_4^5 e^x \, dx = \left[ e^x \right]_4^5 = e^5 - e^4$

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

To solve this, separate the terms in the integral:

$\int_{-1}^1 (x + 1) \, dx = \int_{-1}^1 x \, dx + \int_{-1}^1 1 \, dx$

The first part, $\int_{-1}^1 x \, dx$, is zero because it’s an odd function over a symmetric interval. For the second part:

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

Thus, $\int_{-1}^1 (x + 1) \, dx = 2$.

4. $\int_2^3 \frac{1}{x} \, dx$

The antiderivative of $\frac{1}{x}$ is $\ln |x|$.

$\int_2^3 \frac{1}{x} \, dx = \left[ \ln |x| \right]_2^3 = \ln 3 - \ln 2 = \ln \left( \frac{3}{2} \right)$

5. $\int_1^2 (4x^3 - 5x^2 + 6x + 9) \, dx$

For each term, find the antiderivative:

$\int 4x^3 \, dx = x^4, \quad \int -5x^2 \, dx = -\frac{5x^3}{3}, \quad \int 6x \, dx = 3x^2, \quad \int 9 \, dx = 9x$

So,

$\int_1^2 (4x^3 - 5x^2 + 6x + 9) \, dx = \left[ x^4 - \frac{5x^3}{3} + 3x^2 + 9x \right]_1^2$

Calculate each term at $x = 2$ and $x = 1$:

For $x = 2$:

$2^4 - \frac{5 \cdot 2^3}{3} + 3 \cdot 2^2 + 9 \cdot 2 = 16 - \frac{40}{3} + 12 + 18 = \frac{62}{3}$

For $x = 1$:

$1^4 - \frac{5 \cdot 1^3}{3} + 3 \cdot 1^2 + 9 \cdot 1 = 1 - \frac{5}{3} + 3 + 9 = \frac{20}{3}$

Subtracting:

$\int_1^2 (4x^3 - 5x^2 + 6x + 9) \, dx = \frac{62}{3} - \frac{20}{3} = \frac{42}{3} = 14$

Summary of Results:

1. $\int_2^3 x^2 \, dx = \frac{19}{3}$
2. $\int_4^5 e^x \, dx = e^5 - e^4$
3. $\int_{-1}^1 (x + 1) \, dx = 2$
4. $\int_2^3 \frac{1}{x} \, dx = \ln \left( \frac{3}{2} \right)$
5. $\int_1^2 (4x^3 - 5x^2 + 6x + 9) \, dx = 14$

Would you like further explanations on any of these?

Here are 5 related questions to expand your knowledge:

1. How do you integrate polynomial functions in general?
2. Why does integrating an odd function over a symmetric interval result in zero?
3. What are some applications of the exponential integral in real-world scenarios?
4. How do properties of logarithms simplify integral calculations?
5. How can we apply the Fundamental Theorem of Calculus to check our answers?

Tip: For functions involving symmetry, identifying odd/even properties can simplify integral calculations over symmetric intervals.