The function $F(x)$, where $x$ is a non-negative real number, is the result of subtracting the integer part of $x$ from $x$. For example, $F(3)=0$, $F(5.43)=0.43$, $F(\pi)=0.14159265\ldots$.

We are asked to find an expression for:

$$\int_{0}^{\sqrt{k}} F(x^{2}) \, dx$$ where $k$ is a positive integer.

Solution:

We know that $F(x) = x - \lfloor x \rfloor$. Thus: $$F(x^{2}) = x^{2} - \lfloor x^{2} \rfloor$$ for $0 \leq x \leq \sqrt{k}$.

So the integral becomes: $$\int_{0}^{\sqrt{k}} F(x^{2}) \, dx = \int_{0}^{\sqrt{k}} x^{2} \, dx - \int_{0}^{\sqrt{k}} \lfloor x^{2} \rfloor \, dx$$

The first part is straightforward: $$\int_{0}^{\sqrt{k}} x^{2} \, dx = \frac{1}{3}k^{3/2}.$$

Now consider the second part. For each integer $n = 0,1,2,\dots,k-1$, on the interval $[\sqrt{n}, \sqrt{n+1})$ we have $\lfloor x^{2} \rfloor = n$.

Thus: $$\int_{0}^{\sqrt{k}} \lfloor x^{2} \rfloor \, dx = \sum_{n=0}^{k-1} \int_{\sqrt{n}}^{\sqrt{n+1}} n \, dx = \sum_{n=0}^{k-1} n \left(\sqrt{n+1} - \sqrt{n}\right).$$

Therefore, the required expression is: $$\int_{0}^{\sqrt{k}} F(x^{2}) \, dx = \frac{1}{3}k^{3/2} - \sum_{n=0}^{k-1} n \left(\sqrt{n+1} - \sqrt{n}\right).$$

Answer: (F) $\;\;\frac{1}{3}k^{3/2} + \Big(\sum_{n=0}^{k-1} \sqrt{n}\Big) + (1-k)\sqrt{k}$