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A solution with the floor function incorporated...

Since $\lfloor x \rfloor \, \le \, x \,<\, \lfloor x \rfloor + 1$ for all $x$ , it follows that $x-1 \; < \; \lfloor x \rfloor \; \le \; x$ for all $x$ , and hence that $\tfrac12n(n+1)x - n \; = \; \sum_{j=1}^n (jx - 1) \; < \; \sum_{j=1}^n \lfloor jx \rfloor \; \le \; \sum_{j=1}^n jx \; = \; \tfrac12n(n+1)x$ so that $\tfrac{n+1}{2n}x - \tfrac{1}{n} \; < \; \tfrac{1}{n^2}\sum_{j=1}^n \lfloor jx \rfloor \; \le \; \tfrac{n+1}{2n}x$ which tells us that $f(x) = \tfrac12x$ for all $x$ . The area between $y = f(x)$ and $y = x^2$ is $A \; = \; \int_0^{\frac12} \big(\tfrac12x - x^2\big)\,dx \; = \; \tfrac{1}{48}$ and so the answer is $\int_0^{48} \tfrac12x\,dx \; = \; 24^2 = \boxed{576}$

You can also use Stolz theorem for calculating $f(x) = \frac{x}{2}$ and the rest would be equal to Mark Henning's solution.

Note .- $\lfloor nx \rfloor = nx - \{nx\} \sim nx, \text{ as } n \to \infty$ , where $\{ \cdot \}$ denotes the fractional part function . For calculating $A$ , you need to solve the equation $x^2 = \frac{x}{2} \Rightarrow x = 0, \frac{1}{2}$

By property of $\text{GIF}$ function $\large{x-1 \leq \lfloor x\rfloor < x}$ So we can now express the summation in this form of an inequality.Now we can easily evaluate the limit and it turns out that both limits are equal to be $L = x/2$ .Sandwich Theorem tells us that this is our limit.The Area can be found out using $A = \displaystyle \large \int (f(x) - g(x))dx$ The points of intersection of 2 functions are $(0,0)$ and $(1/2 , 1/4)$ yields $A = 1/48$ .The final answer is thus $\displaystyle \large \int_0^{48} \dfrac{x}{2} dx = \boxed{576}$ .