Let f:[0, infty) rightarrow [0, infty) be def | vedclass

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(C) Given $f(x) = \int_{0}^{x} [y] \, dy$. For $x \in [n, n+1)$,where $n \in \mathbb{N}_0$,we have $[y] = n$ for $y \in [n, x)$. Thus,$f(x) = \int_{0}

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

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(C) Given $f(x) = \int_{0}^{x} [y] \, dy$. For $x \in [n, n+1)$,where $n \in \mathbb{N}_0$,we have $[y] = n$ for $y \in [n, x)$. Thus,$f(x) = \int_{0}^{1} 0 \, dy + \int_{1}^{2} 1 \, dy + \dots + \int_{n-1}^{n} (n-1) \, dy + \int_{n}^{x} n \, dy$. $f(x) = 0 + 1 + 2 + \dots + (n-1) + n(x-n) = \frac{(n-1)n}{2} + nx - n^2$. Since $n = [x]$,we have $f(x) = \frac{[x]([x]-1)}{2} + [x](x-[x])$. At any integer $x=n$,the left-hand limit is $\lim_{x 	o n^-} f(x) = \sum_{k=0}^{n-1} k = \frac{(n-1)n}{2}$ and the right-hand limit is $\lim_{x 	o n^+} f(x) = \frac{(n-1)n}{2} + n(n-n) = \frac{n(n-1)}{2}$. Since the limits are equal,$f(x)$ is continuous for all $x \geq 0$. However,the derivative $f'(x) = [x]$ is discontinuous at integer points because $\lim_{x 	o n^-} f'(x) = n-1$ and $\lim_{x 	o n^+} f'(x) = n$. Therefore,$f$ is continuous everywhere but not differentiable at integer points.

Let $f:[0, \infty) \rightarrow [0, \infty)$ be defined as $f(x) = \int_{0}^{x} [y] \, dy$,where $[x]$ is the greatest integer less than or equal to $x$. Which of the following is true?

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