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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:

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

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### 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? 

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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.

Evaluate the following definite integrals: ∫2^3 x^2 dx, ∫4^5 e^x dx, ∫-1^1 (x + 1) dx, ∫2^3 (1/x) dx, ∫1^2 (4x^3 - 5x^2 + 6x + 9) dx.

Step-by-step solutions to evaluate definite integrals such as ∫2^3 x^2 dx, ∫4^5 e^x dx, ∫-1^1 (x + 1) dx, ∫2^3 (1/x) dx, and ∫1^2 (4x^3 - 5x^2 + 6x + 9) dx. Learn to apply antiderivatives and the Fundamental Theorem of Calculus to solve integrals, covering calculus concepts suitable for Grades 11-12.

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