Why can't I find the area bounded by two curves this way?

Question

I am trying to find the are bounded by:
$x^2+3x-1$ and $x^2+3 \lfloor(x)\rfloor-1$

I first noticed that The area bounded by the curve can be represented as:
$\int_{1}^{2} 3x-3 dx + \int_{2}^{3} 3x-6 dx +...+ \int_{n_1}^{n} 3x+3-3n dx $ Then, we can rewrite the first integral as: $\int_{1}^{2} 3x-6 dx +\int_{1}^{2} 3 dx $.

So, we can combine the first two integrals. If we we peat the process, we can get: $\int_{1}^{n} 3x-3n+3 dx + \frac{n-1}{2}(3n)$.

But end up, it is wrong. Note:
The reason for $\frac{n-1}{2}(3n)$ is that for every two integrals, we would get 3 subtracted out, and the first is $\int_{1}^{2} 3 dx$ the second one is $\int_{1}^{3} 3 dx$, and so on. So we can build a arithmetic series here.
The actual answer of the bounded area is $\frac{3n}{2}+\frac{3}{2}$

Could you help me to check my result and tell me what mistakes did I make?
Also, if you have any quick ways to do this, pls tell me, since I don’t know how to deal with integral of the floor function. I saw some answers that could solve it directly even with floor function.

Thank you so much for your reply

Answer 1

I can't follow exactly what you are saying. The integral is equal to $$\int_{1}^{2} 3x-3 \mathrm{d}x + \int_{2}^{3} 3x-6 \mathrm{d}x +\cdots+ \int_{n-1}^{n} 3x-3(n-1) \mathrm{d}x$$ as you said, but I can't understand exactly what you do next. The easy thing to do is to rewrite is as $$\int_1^n3x\mathrm{d}x-(3+6+\cdots3(n-1))$$ and go from there.

EDIT

In answer to the OP's comment, the sum comes from $$ -\int_1^2 3\mathrm{d}x -\int_2^3 6\mathrm{d}x-\cdots-\int_{n-1}^n 3(n-1)\mathrm{d}x$$ In each case, we have the integral of a constant over an interval of length $1$.