Answer to Question #95334 in Real Analysis for Rajni

Question

Let f: [-5,5]-->R be defined by f(x) =5[x] +x^2 , where [x] denotes the greatest integer function . Show that this function is integrable .Is this function also differentiable ? Justify your answer ?

Answer 1

$\int\limits_{-5}^5f(x)dx=\int\limits_{-5}^5x^2dx+5\int\limits_{-5}^5[x]dx$. The integral $\int\limits_{-5}^5x^2dx$ exists. Show that integral $\int\limits_{-5}^5[x]dx$ also exists.

$\int\limits_{-5}^5[x]dx=\sum\limits_{k=-5}^4\int\limits_k^{k+1}[x]dx=\sum\limits_{k=-5}^4\int\limits_k^{k+1}kdx$ , so integral $\int\limits_{-5}^5[x]dx$ exists.

$f$ is not differentiable in $[-4,5]\cap\mathbb Z$, because left derivative of $f$ in this set do not exist .

Indeed, let $k\in[-4,5]\cap\mathbb Z$ , so left derivative in point $k$ is$\lim\limits_{x\to 0+}\frac{f(k)-f(k-x)}{x}=\lim\limits_{x\to 0+}\frac{5[k]+k^2-5[k-x]-(k-x)^2}{x}=$

$=\lim\limits_{x\to 0+}\frac{5k+k^2-5[k-x]-k^2+2kx-x^2}{x}=\lim\limits_{x\to 0+}\frac{5k-5[k-x]+2kx-x^2}{x}=$

$=2k+5\lim\limits_{x\to 0+}\frac{k-[k-x]}{x}$

If $x\in (0,1)$ , then $[k-x]=k-1$ , so $\lim\limits_{x\to 0+}\frac{k-[k-x]}{x}=\lim\limits_{x\to 0+}\frac{k-(k-1)}{x}=+\infty$

So we obtain that left derivatives of $f$ in $[-4,5]\cap\mathbb Z$ equal $+\infty$ , that is, do not exist.

Answer: $f$ is integrable, but not differentiable.