The problem asks us to evaluate the definite integral of $x^2 - 3x$ from 1 to 2, and subtract from that the definite integral of $x^2 - 4x - 1$ from 1 to 2. In other words, evaluate $\int_1^2 (x^2 - 3x) \, dx - \int_1^2 (x^2 - 4x - 1) \, dx$.

1. Problem Description

The problem asks us to evaluate the definite integral of $x^2 - 3x$ from 1 to 2, and subtract from that the definite integral of $x^2 - 4x - 1$ from 1 to

2. In other words, evaluate

$\int_1^2 (x^2 - 3x) \, dx - \int_1^2 (x^2 - 4x - 1) \, dx$.

2. Solution Steps

We can combine the two integrals into one since they have the same limits of integration.

$\int_1^2 (x^2 - 3x) \, dx - \int_1^2 (x^2 - 4x - 1) \, dx = \int_1^2 [(x^2 - 3x) - (x^2 - 4x - 1)] \, dx$

Simplifying the expression inside the integral:

$\int_1^2 (x^2 - 3x - x^2 + 4x + 1) \, dx = \int_1^2 (x + 1) \, dx$

Now, we can integrate:

$\int_1^2 (x + 1) \, dx = \left[ \frac{x^2}{2} + x \right]_1^2$

Evaluating the antiderivative at the upper and lower limits:

$\left( \frac{2^2}{2} + 2 \right) - \left( \frac{1^2}{2} + 1 \right) = \left( \frac{4}{2} + 2 \right) - \left( \frac{1}{2} + 1 \right) = (2 + 2) - \left( \frac{1}{2} + \frac{2}{2} \right) = 4 - \frac{3}{2} = \frac{8}{2} - \frac{3}{2} = \frac{5}{2}$

3. Final Answer

$\frac{5}{2}$