2
$\begingroup$

I want to calculate the following integral $$\int_{0}^{20}\{x\}^{2}+\left \lfloor{x}\right \rfloor^{2}dx $$

We have that $x=\{x\}+\left \lfloor{x}\right \rfloor$, then $x^2-2\{x\}\left \lfloor{x}\right \rfloor =\{x\}^{2}+\left \lfloor{x}\right \rfloor^{2}$

$$\int_{0}^{20}\{x\}^{2}+\left \lfloor{x}\right \rfloor^{2}dx=\int_{0}^{20}x^2-2\{x\}\left \lfloor{x}\right \rfloor dx= \frac{20^{3}}{3}-2\int_{0}^{20}\{x\}\left \lfloor{x}\right \rfloor dx$$

How can I find the value of $\int_{0}^{20}\{x\}\left \lfloor{x}\right \rfloor dx$?

$\endgroup$
1
$\begingroup$

Hint

The main idea here is that in each interval of length $1$, the fractional part of $x$ behaves exactly like $x$ (without the integer increase). The greatest integer part over each interval is a constant and is equal to the lower bound of that interval. Now integrating a constant from $5$ to $6$ is same as integrating it from $0$ to $1$.

$$\int_0^{20}(\{x\}^{2}+\left \lfloor{x}\right \rfloor^2) \, dx $$

$$=20\int_0^1 x^2 \, dx + \int_0^1 (0^2 + 1^2 + 2^2 + \cdots + 19^2) \, dx $$

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Why does the second integral have a factor of $20$? $\qquad$ $\endgroup$ – Michael Hardy Mar 11 '16 at 13:37
  • $\begingroup$ $$ \int_0^{20} \lfloor x \rfloor^2 \, dx = \sum_{n=0}^{19} \int_n^{n+1} \lfloor x \rfloor^2 \, dx = \sum_{n=0}^{19} n^2 \ne 20\int_0^1 \sum_{n=0}^{19} n^2\,dx.$$ $\endgroup$ – Michael Hardy Mar 11 '16 at 14:31
  • $\begingroup$ @MichaelHardy You are absolutely right. I have edited and corrected my answer. Thank you so much for your time in correcting an older answer. $\endgroup$ – Shailesh Mar 12 '16 at 8:35
4
$\begingroup$

I think it is easier to proceed in the following way: \begin{align} \int_0^{20}\{x\}^2+\lfloor x\rfloor^2dx & = \sum_{k=0}^{19}\int_k^{k+1}\{x\}^2+\lfloor x\rfloor^2dx\\ & = \sum_{k=0}^{19}k^2 + \int_0^1x^2dx\\ & = \frac{19\cdot20\cdot39}{6} +\frac{20}{3}\\ & = \frac{14840}{3}. \end{align} Of course, from this you can also deduce the value of $\int_0^{20}\{x\}\lfloor x\rfloor dx$.

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

Hint, break it up into $\int_{n-1}^n$ for $n=1,2,3,\dots, 20$ and sum. The functions are much easier on these intervals.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Do you know of any general method for indefinite or improper integrals? $\endgroup$ – ThisIsNotAnId Feb 19 '16 at 23:43
  • $\begingroup$ This is neither an indefinite nor an improper integral. @ThisIsNotAnId $\endgroup$ – Thomas Andrews Feb 19 '16 at 23:44
  • $\begingroup$ I know; I was just curious if there existed any. $\endgroup$ – ThisIsNotAnId Feb 19 '16 at 23:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.