The given integral is:

$I = \int_{6}^{3} (-2x + 1) \log(2x) \, dx$

We will solve this step by step. Since the bounds are reversed ($6$ to $3$), we start by swapping the limits of integration, introducing a negative sign:

$I = -\int_{3}^{6} (-2x + 1) \log(2x) \, dx$

Now let us evaluate the integral systematically.

Step 1: Split the integral

The integrand can be expanded into two terms: $\int (-2x + 1) \log(2x) \, dx = \int -2x \log(2x) \, dx + \int \log(2x) \, dx$

So, we have: $I = -\left[ \int_{3}^{6} -2x \log(2x) \, dx + \int_{3}^{6} \log(2x) \, dx \right]$

Step 2: Handle the individual integrals

1. First integral: $\int -2x \log(2x) \, dx$
   Use integration by parts with $u = \log(2x)$ and $dv = -2x \, dx$.

2. Second integral: $\int \log(2x) \, dx$
   Similarly, use integration by parts with $u = \log(2x)$ and $dv = dx$.

Let me calculate this fully. sympy as sp

Define variables

x = sp.symbols('x') a, b = 3, 6

Define the function

integrand = (-2 * x + 1) * sp.log(2 * x)

Perform the integral with reversed limits (introducing a negative sign)

integral_result = -sp.integrate(integrand, (x, a, b)) integral_resultThe solution to the integral is:

$I = -6\log(6) - \frac{21}{2} + 30\log(12)$

Would you like me to simplify this further or convert it to a numerical value?

Here are 5 related questions to expand your understanding:

1. How does the method of integration by parts work, and when is it applied?
2. What happens when the integration bounds are swapped?
3. Why is the logarithmic term $\log(2x)$ treated separately in the integral?
4. Can this integral be evaluated numerically for approximation?
5. How would the solution change if the limits of integration were positive integers?

Tip: Always simplify logarithmic expressions like $\log(ab)$ as $\log(a) + \log(b)$ to make calculations easier!