The problem asks us to evaluate the definite integral of the polynomial $x^2 - 3x + 5$ from $2$ to $3$. In other words, we need to find $\int_{2}^{3} (x^2 - 3x + 5) \, dx$.

1. Problem Description

The problem asks us to evaluate the definite integral of the polynomial $x^2 - 3x + 5$ from $2$ to $3$. In other words, we need to find $\int_{2}^{3} (x^2 - 3x + 5) \, dx$.

2. Solution Steps

First, we find the indefinite integral of the polynomial:

$\int (x^2 - 3x + 5) \, dx = \int x^2 \, dx - \int 3x \, dx + \int 5 \, dx$

Using the power rule for integration, $\int x^n \, dx = \frac{x^{n+1}}{n+1} + C$, we have:

$\int x^2 \, dx = \frac{x^3}{3} + C_1$

$\int 3x \, dx = 3 \int x \, dx = 3 \cdot \frac{x^2}{2} + C_2 = \frac{3x^2}{2} + C_2$

$\int 5 \, dx = 5x + C_3$

Combining these, we get the indefinite integral:

$\int (x^2 - 3x + 5) \, dx = \frac{x^3}{3} - \frac{3x^2}{2} + 5x + C$

Now, we evaluate the definite integral from 2 to 3:

$\int_{2}^{3} (x^2 - 3x + 5) \, dx = \left[ \frac{x^3}{3} - \frac{3x^2}{2} + 5x \right]_{2}^{3}$

We plug in the upper limit (3) and the lower limit (2) into the expression and subtract:

$= \left( \frac{3^3}{3} - \frac{3(3^2)}{2} + 5(3) \right) - \left( \frac{2^3}{3} - \frac{3(2^2)}{2} + 5(2) \right)$

$= \left( \frac{27}{3} - \frac{27}{2} + 15 \right) - \left( \frac{8}{3} - \frac{12}{2} + 10 \right)$

$= \left( 9 - \frac{27}{2} + 15 \right) - \left( \frac{8}{3} - 6 + 10 \right)$

$= \left( 24 - \frac{27}{2} \right) - \left( \frac{8}{3} + 4 \right)$

$= 24 - \frac{27}{2} - \frac{8}{3} - 4$

$= 20 - \frac{27}{2} - \frac{8}{3}$

$= 20 - \frac{27 \cdot 3}{2 \cdot 3} - \frac{8 \cdot 2}{3 \cdot 2}$

$= 20 - \frac{81}{6} - \frac{16}{6}$

$= 20 - \frac{97}{6}$

$= \frac{120}{6} - \frac{97}{6}$

$= \frac{23}{6}$

3. Final Answer

The final answer is $\frac{23}{6}$.