Integral using the definition of integral is returning double the real answer? I'm trying to evaluate an integral using the definition of the definite integral but for some reason my answer is 12 when it should be 6. I assumed that I'd missed a 2 somewhere that would go in the denominator but I've tried this exercise multiple times and I keep getting the same answer.
My steps:
$$ \int_0^3 (2x-1)dx = \lim_{n\to \infty} \sum_{k=1}^n f\left(\frac{k(3-0)}{n}\right) \frac{3}{n} $$
First I solve the function that is inside the summation.
$$ f\left(\frac{k(3-0)}{n}\right) = f\left(\frac{3k}{n}\right) = 2\left(\frac{3k}{n}\right) - 1 = \frac{6k}{n} - 1 $$
Then I solve the summation.
$$ \sum_{k=1}^{n} \left( \frac{6k}{n} - 1 \right) \frac{3}{n} = \frac{3}{n}\sum_{k=1}^{n} \left( \frac{6k}{n} - 1 \right) = \frac{3}{n}\left(\sum_{k=1}^{n} \frac{6k}{n} - \sum_{k=1}^{n} (-1) \right) = \frac{3}{n}\left( \left(\sum_{k=1}^{n} \frac{6}{n}k \right) - (-n) \right) = \frac{3}{n}\left(\frac{6}{n}\left(\sum_{k=1}^{n}k \right) + n \right) = \frac{3}{n} \left( \frac{6}{n} \cdot \frac{n(n+1)}{2} + n \right) = \frac{3}{n} \left( 3(n+1) + n \right) = \frac{3}{n}(4n + 3) = \frac{12n + 9}{n} $$
Then I find the limit of the solved summation as $n \to \infty$.
$$ \lim_{n\to\infty} \frac{12n + 9}{n} = 12 $$
However, the answer should be 6. What am I doing wrong here?

As Michael said, it's just a sign error:
$$\cdots= \frac{3}{n} \left( \frac{6}{n} \cdot \frac{n(n+1)}{2} \color{red}{-} n \right) = \frac{3}{n} \left( 3(n+1) \color{red}{-} n \right) = \frac{3}{n}(\color{red}{2}n + 3) = \frac{\color{red}{6}n + 9}{n}$$
