Do either $(1)$ or $(2)$:

$(1)$   Prove that $\int_{1}^{k} [x] f'(x) dx = [k] f(k) - \sum_{1}{[k]} f(n),$ where $k > 1,$ and $[z]$ denotes the greatest integer $\leq z.$ Find a similar expression for:   $\int_{1}^{k} [x^2] f'(x) dx.$

$(2)$   A particle moves freely in a straight line except for a resistive force proportional to its speed. Its speed falls from $1,000 \dfrac{ft}{s}$ to $900 \dfrac{ft}{s}$ over $1200 ft.$ Find the time taken to the nearest $0.01 s.$ [No calculators or log tables allowed!]
1. **Prove that** \(\int_{1}^{k} [x] f'(x) \, dx = [k] f(k) - \sum_{n=1}^{[k]} f(n)\), **where** \(k > 1\), **and** \([z]\) **denotes the greatest integer** \(\leq z\).

   For \(k \geq 2\):
   \[
   \int_1^k \lfloor x \rfloor f'(x) \, dx = \sum_{r=1}^{\lfloor k \rfloor - 1} \int_r^{r+1} \lfloor x \rfloor f'(x) \, dx + \int_{\lfloor k \rfloor}^k \lfloor x \rfloor f'(x) \, dx
   \]
   Since \(\lfloor x \rfloor = r\) for \(x \in [r, r+1)\) and \(\lfloor x \rfloor = \lfloor k \rfloor\) for \(x \in [\lfloor k \rfloor, k]\), we have:
   \[
   = \sum_{r=1}^{\lfloor k \rfloor - 1} \int_r^{r+1} r f'(x) \, dx + \int_{\lfloor k \rfloor}^k \lfloor k \rfloor f'(x) \, dx
   \]
   Using the Fundamental Theorem of Calculus:
   \[
   = \sum_{r=1}^{\lfloor k \rfloor - 1} r (f(r+1) - f(r)) + \lfloor k \rfloor (f(k) - f(\lfloor k \rfloor))
   \]
   Distributing and rearranging terms:
   \[
   = \sum_{r=1}^{\lfloor k \rfloor - 1} (r f(r+1) - r f(r)) + \lfloor k \rfloor f(k) - \lfloor k \rfloor f(\lfloor k \rfloor)
   \]
   \[
   = \sum_{r=1}^{\lfloor k \rfloor - 1} (r f(r+1) - r f(r)) + \lfloor k \rfloor f(k) - \lfloor k \rfloor f(\lfloor k \rfloor)
   \]
   \[
   = \sum_{r=1}^{\lfloor k \rfloor - 1} (r f(r+1) - r f(r)) + \lfloor k \rfloor f(k) - \lfloor k \rfloor f(\lfloor k \rfloor)
   \]
   \[
   = \lfloor k \rfloor f(k) - \sum_{r=1}^{\lfloor k \rfloor} f(r)
   \]
   For \(1 < k < 2\):
   \[
   \int_1^k \lfloor x \rfloor f'(x) \, dx = \int_1^k f'(x) \, dx = f(k) - f(1) = \lfloor k \rfloor f(k) - f(1)
   \]
   Summarizing, for \(k > 1\):
   \[
   \boxed{\int_1^k \lfloor x \rfloor f'(x) \, dx = \lfloor k \rfloor f(k) - \sum_{r=1}^{\lfloor k \rfloor} f(r)}
   \]

2. **Find a similar expression for** \(\int_{1}^{k} [x^2] f'(x) \, dx\).

   For \(k \geq \sqrt{2}\):
   \[
   \int_1^k \lfloor x^2 \rfloor f'(x) \, dx = \sum_{r=1}^{\lfloor k^2 \rfloor - 1} \int_{\sqrt{r}}^{\sqrt{r+1}} \lfloor x^2 \rfloor f'(x) \, dx + \int_{\sqrt{\lfloor k^2 \rfloor}}^k \lfloor x^2 \rfloor f'(x) \, dx
   \]
   Since \(\lfloor x^2 \rfloor = r\) for \(x \in [\sqrt{r}, \sqrt{r+1})\) and \(\lfloor x^2 \rfloor = \lfloor k^2 \rfloor\) for \(x \in [\sqrt{\lfloor k^2 \rfloor}, k]\), we have:
   \[
   = \sum_{r=1}^{\lfloor k^2 \rfloor - 1} \int_{\sqrt{r}}^{\sqrt{r+1}} r f'(x) \, dx + \int_{\sqrt{\lfloor k^2 \rfloor}}^k \lfloor k^2 \rfloor f'(x) \, dx
   \]
   Using the Fundamental Theorem of Calculus:
   \[
   = \sum_{r=1}^{\lfloor k^2 \rfloor - 1} r (f(\sqrt{r+1}) - f(\sqrt{r})) + \lfloor k^2 \rfloor (f(k) - f(\sqrt{\lfloor k^2 \rfloor}))
   \]
   Distributing and rearranging terms:
   \[
   = \sum_{r=1}^{\lfloor k^2 \rfloor - 1} (r f(\sqrt{r+1}) - r f(\sqrt{r})) + \lfloor k^2 \rfloor f(k) - \lfloor k^2 \rfloor f(\sqrt{\lfloor k^2 \rfloor})
   \]
   \[
   = \lfloor k^2 \rfloor f(k) - \sum_{r=1}^{\lfloor k^2 \rfloor} f(\sqrt{r})
   \]
   For \(1 < k < \sqrt{2}\):
   \[
   \int_1^k \lfloor x^2 \rfloor f'(x) \, dx = \int_1^k f'(x) \, dx = f(k) - f(1) = \lfloor k^2 \rfloor f(k) - f(\sqrt{1})
   \]
   Summarizing, for \(k > 1\):
   \[
   \boxed{\int_1^k \lfloor x^2 \rfloor f'(x) \, dx = \lfloor k^2 \rfloor f(k) - \sum_{r=1}^{\lfloor k^2 \rfloor} f(\sqrt{r})}
   \]