The problem is to evaluate the definite integrals $\int_{1}^{2} (4x^2+3x-1)dx - \int_{1}^{3} (4x^2+3x-1)dx$.

1. Problem Description

The problem is to evaluate the definite integrals $\int_{1}^{2} (4x^2+3x-1)dx - \int_{1}^{3} (4x^2+3x-1)dx$.

2. Solution Steps

First, we find the indefinite integral of the polynomial $4x^2 + 3x - 1$. Using the power rule for integration, we have

$\int (4x^2 + 3x - 1)dx = \frac{4}{3}x^3 + \frac{3}{2}x^2 - x + C$.

Let $F(x) = \frac{4}{3}x^3 + \frac{3}{2}x^2 - x$. Then,

$\int_{1}^{2} (4x^2+3x-1)dx = F(2) - F(1) = (\frac{4}{3}(2^3) + \frac{3}{2}(2^2) - 2) - (\frac{4}{3}(1^3) + \frac{3}{2}(1^2) - 1) = (\frac{32}{3} + 6 - 2) - (\frac{4}{3} + \frac{3}{2} - 1) = \frac{32}{3} + 4 - \frac{4}{3} - \frac{3}{2} + 1 = \frac{28}{3} + 5 - \frac{3}{2} = \frac{56}{6} + \frac{30}{6} - \frac{9}{6} = \frac{77}{6}$.

$\int_{1}^{3} (4x^2+3x-1)dx = F(3) - F(1) = (\frac{4}{3}(3^3) + \frac{3}{2}(3^2) - 3) - (\frac{4}{3}(1^3) + \frac{3}{2}(1^2) - 1) = (\frac{4}{3}(27) + \frac{3}{2}(9) - 3) - (\frac{4}{3} + \frac{3}{2} - 1) = (36 + \frac{27}{2} - 3) - (\frac{4}{3} + \frac{3}{2} - 1) = 33 + \frac{27}{2} - \frac{4}{3} - \frac{3}{2} + 1 = 34 + \frac{24}{2} - \frac{4}{3} = 34 + 12 - \frac{4}{3} = 46 - \frac{4}{3} = \frac{138}{3} - \frac{4}{3} = \frac{134}{3}$.

Therefore, $\int_{1}^{2} (4x^2+3x-1)dx - \int_{1}^{3} (4x^2+3x-1)dx = \frac{77}{6} - \frac{134}{3} = \frac{77}{6} - \frac{268}{6} = \frac{-191}{6}$.

3. Final Answer

-191/6