Key Concepts and Formulas

• Greatest Integer Function, $[x]$: This function is piecewise constant. For any integer $n$, $[x] = n$ for $x \in [n, n+1)$. It has jump discontinuities at every integer.
• Continuity of a function $f$ at a point $c$: $\lim_{x \to c^-} f(x) = \lim_{x \to c^+} f(x) = f(c)$.
• Differentiability of a function $f$ at a point $c$: The left-hand derivative $f'(c^-) = \lim_{h \to 0^-} \frac{f(c+h) - f(c)}{h}$ and the right-hand derivative $f'(c^+) = \lim_{h \to 0^+} \frac{f(c+h) - f(c)}{h}$ must exist and be equal.
• Fundamental Theorem of Calculus (FTC) Part 1: If $F(x) = \int_a^x g(t) dt$, then $F'(x) = g(x)$ if $g(x)$ is continuous at $x$. If $g(x)$ is discontinuous at $x$, we need to evaluate the derivative using limits.

Step-by-Step Solution

Step 1: Evaluate the definite integral $f(x)$ We need to understand the behavior of $f(x) = \int_0^x {[y]dy}$ for different values of $x$. The integrand $[y]$ changes its value at integer points. Let's consider intervals.

For $x \in [0, 1)$: $f(x) = \int_0^x {[y]dy} = \int_0^x 0 \, dy = 0$.

For $x \in [1, 2)$: $f(x) = \int_0^x {[y]dy} = \int_0^1 {[y]dy} + \int_1^x {[y]dy} = \int_0^1 0 \, dy + \int_1^x 1 \, dy = 0 + [y]_1^x = x - 1$.

For $x \in [2, 3)$: $f(x) = \int_0^x {[y]dy} = \int_0^1 {[y]dy} + \int_1^2 {[y]dy} + \int_2^x {[y]dy} = \int_0^1 0 \, dy + \int_1^2 1 \, dy + \int_2^x 2 \, dy = 0 + (2-1) + [2y]_2^x = 1 + 2x - 4 = 2x - 3$.

In general, for $x \in [n, n+1)$ where $n$ is a non-negative integer: $f(x) = \int_0^x {[y]dy} = \sum_{k=0}^{n-1} \int_k^{k+1} {[y]dy} + \int_n^x {[y]dy}$ $f(x) = \sum_{k=0}^{n-1} \int_k^{k+1} k \, dy + \int_n^x n \, dy$ $f(x) = \sum_{k=0}^{n-1} k(k+1-k) + n(x-n)$ $f(x) = \sum_{k=0}^{n-1} k + n(x-n)$ $f(x) = \frac{(n-1)n}{2} + nx - n^2$ $f(x) = nx - \frac{n(n+1)}{2}$.

So, the function $f(x)$ can be written piecewise as: $f(x) = \begin{cases} 0 & \text{if } 0 \le x < 1 \\ x-1 & \text{if } 1 \le x < 2 \\ 2x-3 & \text{if } 2 \le x < 3 \\ 3x - 6 & \text{if } 3 \le x < 4 \\ \vdots & \\ nx - \frac{n(n+1)}{2} & \text{if } n \le x < n+1 \end{cases}$

Step 2: Analyze the continuity of $f(x)$ We need to check continuity at integer points. Let's check at an arbitrary integer $n \ge 1$. We need to compare $\lim_{x \to n^-} f(x)$, $\lim_{x \to n^+} f(x)$, and $f(n)$.

From Step 1, for $x \in [n-1, n)$: $f(x) = (n-1)x - \frac{(n-1)n}{2}$. So, $\lim_{x \to n^-} f(x) = (n-1)n - \frac{(n-1)n}{2} = \frac{(n-1)n}{2}$.

For $x \in [n, n+1)$: $f(x) = nx - \frac{n(n+1)}{2}$. So, $\lim_{x \to n^+} f(x) = n \cdot n - \frac{n(n+1)}{2} = n^2 - \frac{n^2+n}{2} = \frac{2n^2 - n^2 - n}{2} = \frac{n^2-n}{2} = \frac{n(n-1)}{2}$.

Also, $f(n)$ is calculated using the formula for $x \in [n, n+1)$, so $f(n) = n \cdot n - \frac{n(n+1)}{2} = \frac{n(n-1)}{2}$.

Since $\lim_{x \to n^-} f(x) = \lim_{x \to n^+} f(x) = f(n) = \frac{n(n-1)}{2}$, the function $f(x)$ is continuous at every integer point $n \ge 1$. For $x=0$, $f(0) = 0$. For $x \in (0, 1)$, $f(x)=0$. So $f$ is continuous at $x=0$. Therefore, $f(x)$ is continuous at every point in $[0, \infty)$.

Step 3: Analyze the differentiability of $f(x)$ We need to check differentiability at integer points. Let's check at an arbitrary integer $n \ge 1$. We will calculate the left-hand derivative and the right-hand derivative at $x=n$.

Left-hand derivative at $x=n$: $f'(n^-) = \lim_{h \to 0^-} \frac{f(n+h) - f(n)}{h}$. For $h < 0$ and small, $n+h$ is in the interval $[n-1, n)$. So, $f(n+h) = (n-1)(n+h) - \frac{(n-1)n}{2}$. And $f(n) = n \cdot n - \frac{n(n+1)}{2} = \frac{n(n-1)}{2}$. $f'(n^-) = \lim_{h \to 0^-} \frac{(n-1)(n+h) - \frac{(n-1)n}{2} - \frac{n(n-1)}{2}}{h}$ $f'(n^-) = \lim_{h \to 0^-} \frac{(n-1)n + (n-1)h - \frac{n^2-n+n^2-n}{2}}{h}$ $f'(n^-) = \lim_{h \to 0^-} \frac{(n-1)n + (n-1)h - \frac{2n^2-2n}{2}}{h}$ $f'(n^-) = \lim_{h \to 0^-} \frac{(n-1)n + (n-1)h - (n^2-n)}{h}$ $f'(n^-) = \lim_{h \to 0^-} \frac{n^2-n + (n-1)h - n^2+n}{h} = \lim_{h \to 0^-} \frac{(n-1)h}{h} = n-1$.

Right-hand derivative at $x=n$: $f'(n^+) = \lim_{h \to 0^+} \frac{f(n+h) - f(n)}{h}$. For $h > 0$ and small, $n+h$ is in the interval $[n, n+1)$. So, $f(n+h) = n(n+h) - \frac{n(n+1)}{2}$. And $f(n) = \frac{n(n-1)}{2}$. $f'(n^+) = \lim_{h \to 0^+} \frac{n(n+h) - \frac{n(n+1)}{2} - \frac{n(n-1)}{2}}{h}$ $f'(n^+) = \lim_{h \to 0^+} \frac{n^2 + nh - \frac{n^2+n+n^2-n}{2}}{h}$ $f'(n^+) = \lim_{h \to 0^+} \frac{n^2 + nh - \frac{2n^2}{2}}{h}$ $f'(n^+) = \lim_{h \to 0^+} \frac{n^2 + nh - n^2}{h} = \lim_{h \to 0^+} \frac{nh}{h} = n$.

Since $f'(n^-) = n-1$ and $f'(n^+) = n$, and $n-1 \neq n$ for any integer $n$, the function $f(x)$ is not differentiable at any integer point $n \ge 1$. For $x=0$, $f(0)=0$. For $x \in (0,1)$, $f(x)=0$. The right-hand derivative at $0$ is $f'(0^+) = \lim_{h \to 0^+} \frac{f(h)-f(0)}{h} = \lim_{h \to 0^+} \frac{0-0}{h} = 0$. The left-hand derivative is not applicable as the domain starts at 0. However, since the function is constant on $[0,1)$, it is differentiable on $(0,1)$ with derivative 0.

For $x$ not an integer, say $x \in (n, n+1)$ for some integer $n \ge 0$. In this interval, $f(x) = nx - \frac{n(n+1)}{2}$. The derivative is $f'(x) = n$. Since the derivative is a constant value $n$ on each open interval $(n, n+1)$, $f(x)$ is differentiable at every non-integer point.

Therefore, $f(x)$ is continuous at every point in $[0,\infty)$ and differentiable except at the integer points.

Step 4: Evaluate the options (A) $f$ is continuous at every point in $[0,\infty)$ and differentiable except at the integer points. This matches our findings. (B) $f$ is both continuous and differentiable except at the integer points in $[0,\infty)$. This is incorrect because $f$ is continuous at integer points. (C) $f$ is continuous everywhere except at the integer points in $[0,\infty)$. This is incorrect because $f$ is continuous at integer points. (D) $f$ is differentiable at every point in $[0,\infty)$. This is incorrect because $f$ is not differentiable at integer points.

Common Mistakes & Tips

• Misinterpreting the FTC: The FTC part 1 states $F'(x) = g(x)$ if $g(x)$ is continuous. Here, the integrand $[y]$ is discontinuous at integers, so we cannot directly apply the FTC to find the derivative at integer points. We must use the definition of the derivative.
• Confusing continuity and differentiability: A function can be continuous at a point but not differentiable there (e.g., $|x|$ at $x=0$). In this case, $f(x)$ is continuous at integers but not differentiable.
• Careful evaluation of piecewise functions: When dealing with piecewise functions, especially those defined by integrals of step functions, it is crucial to correctly set up the integrals for each interval and then check the behavior at the boundaries.

Summary

We first evaluated the definite integral $f(x) = \int_0^x {[y]dy}$ by considering intervals based on the greatest integer function. This resulted in a piecewise linear function. We then analyzed the continuity of $f(x)$ at integer points by comparing the left-hand limit, right-hand limit, and function value. We found that $f(x)$ is continuous everywhere. Next, we analyzed the differentiability at integer points by calculating the left-hand and right-hand derivatives and found they were not equal, meaning $f(x)$ is not differentiable at integer points. For non-integer points, $f(x)$ is differentiable. This leads to the conclusion that $f$ is continuous at every point in $[0,\infty)$ and differentiable except at the integer points.

The final answer is \boxed{A}.