Let $$f: [0, \infty) \to [0, \infty)$$ 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?

We are given the function $$f : [0,\infty) \to [0,\infty)$$ defined by the rule

$$f(x)=\int_{0}^{x}[y]\,dy,$$

where $$[y]$$ denotes the greatest integer less than or equal to $$y$$ (the “floor-function”). Our task is to test the four statements about the continuity and differentiability of $$f$$.

First we must understand the exact algebraic form of $$f(x)$$ on every sub-interval between two consecutive integers. Let us take an arbitrary non-negative integer $$n$$ and look at all $$x$$ that satisfy

$$n\le x<n+1.$$

For any such $$x$$ we always have $$[y]=n$$ whenever the variable of integration $$y$$ itself satisfies $$n\le y<n+1$$. Hence, over the interval $$[n,n+1)$$, the integrand is the constant $$n$$. To integrate from 0 to $$x$$ we split the whole range $$[0,x]$$ into unit pieces:

$$ \int_{0}^{x}[y]\,dy =\int_{0}^{1}[y]\,dy +\int_{1}^{2}[y]\,dy +\cdots +\int_{n-1}^{n}[y]\,dy +\int_{n}^{x}[y]\,dy. $$

On each earlier unit interval $$[k,k+1]$$ (where $$k=0,1,\dots ,n-1$$) the integrand equals the integer $$k$$. Relying on the fact that the length of every such interval is exactly 1, we evaluate them one by one:

$$ \int_{k}^{k+1}[y]\,dy=\int_{k}^{k+1}k\,dy =k\int_{k}^{k+1}dy =k(1)=k. $$

Hence the sum of the first $$n$$ integrals is

$$ \sum_{k=0}^{n-1}k =0+1+2+\dots +(n-1). $$

We recall the formula for the sum of the first $$m$$ natural numbers:

$$ 0+1+2+\dots +m=\frac{m(m+1)}{2}. $$

Using $$m=n-1$$ we get

$$ \sum_{k=0}^{n-1}k=\frac{(n-1)n}{2}. $$

The last part of the integral is over $$[n,x]$$ where the integrand is the constant $$n$$. Its value is

$$ \int_{n}^{x}n\,dy =n\int_{n}^{x}dy =n(x-n). $$

Combining everything, for every $$x$$ with $$n\le x<n+1$$ we have

$$ f(x)=\frac{(n-1)n}{2}+n(x-n). $$

That is an explicit linear expression in $$x$$ inside each open interval $$(n,n+1).$$ Now we test continuity at the integer point $$x=n.$$ We must compare the left-hand limit, the value actually taken, and the right-hand limit.

Left-hand limit as $$x\to n^-$$ (use the formula with $$n-1$$ because $$x$$ lies in $$(n-1,n)$$):

$$ \lim_{x\to n^-}f(x) =\frac{(n-2)(n-1)}{2}+(n-1)\bigl(n-(n-1)\bigr) =\frac{(n-2)(n-1)}{2}+(n-1)(1). $$

Simplifying term by term,

$$ (n-1)(1)=n-1, $$

so

$$ \frac{(n-2)(n-1)}{2}+n-1 =\frac{(n-1)\bigl(n-2+2\bigr)}{2} =\frac{(n-1)n}{2}. $$

Now the actual value when $$x=n$$ is obtained by direct substitution in the defining integral:

$$ f(n)=\int_{0}^{n}[y]\,dy. $$

Everything up to $$y=n$$ has been covered earlier, hence

$$ f(n)=\frac{(n-1)n}{2}, $$

the same number that just appeared.

Right-hand limit as $$x\to n^+$$ (use the formula with $$n$$ because now $$x$$ lies in $$(n,n+1)$$):

$$ \lim_{x\to n^+}f(x) =\frac{(n-1)n}{2}+n\bigl(n-n\bigr) =\frac{(n-1)n}{2}+n\cdot0 =\frac{(n-1)n}{2}. $$

Therefore the left-hand limit, the value at $$x=n$$, and the right-hand limit are identical. Hence $$f$$ is continuous at every integer, and of course it is linear (hence continuous) on every open interval between integers. All together, $$f$$ is continuous on the whole half-line $$[0,\infty).$$

Now we turn to differentiability. Whenever $$x$$ lies strictly between two integers, say $$n<x<n+1,$$ the integrand $$[y]$$ is continuous at $$y=x$$, so the Fundamental Theorem of Calculus (stated as “If $$F(x)=\int_{a}^{x}g(t)dt$$ and $$g$$ is continuous at $$x$$ then $$F'(x)=g(x)$$”) applies. Hence

$$ f'(x)=[x]=n \quad\text{for }n<x<n+1. $$

Thus $$f'(x)$$ exists and equals a constant integer on every open interval between integers.

Finally, examine differentiability at the integer point $$x=n$$ itself. The derivative, if it exists, must match the two one-sided limits of the difference quotient. We compute:

Left-hand derivative

$$ \lim_{h\to0^-}\frac{f(n+h)-f(n)}{h} =\lim_{h\to0^-}\frac{(n-1)(n-2)/2+(n-1)(h+1)-\frac{(n-1)n}{2}}{h} =\lim_{h\to0^-}\frac{(n-1)h}{h}=n-1. $$

Right-hand derivative

$$ \lim_{h\to0^+}\frac{f(n+h)-f(n)}{h} =\lim_{h\to0^+}\frac{\frac{(n-1)n}{2}+n\,h-\frac{(n-1)n}{2}}{h} =\lim_{h\to0^+}\frac{n\,h}{h}=n. $$

Because $$n-1\ne n,$$ the two one-sided derivatives are unequal, so $$f'(n)$$ does not exist. Therefore $$f$$ is not differentiable at any integer, but it is differentiable everywhere else.

Gathering all conclusions:

• $$f$$ is continuous at every point of $$[0,\infty).$$
• $$f$$ fails to be differentiable precisely at the integers and is differentiable everywhere else.

This matches exactly the wording of Option A.

Hence, the correct answer is Option A.