5
$\begingroup$

I need help for solving this demostration, I appreciate your suggestions very much.

$$\begin{array}{rclr} \int ^{n}_{0}[x] dx= \frac{n(n-1)}{2} \end{array}$$

Pd. If you have any suggestion of a book that deepens on the subject please communicate it.

$\endgroup$
  • 2
    $\begingroup$ @KennyLog_ins I don't agree with your edit. Perhaps surprisingly, integration by parts means something completely different. The meaning of the original phrase "interger part integration" in the title referred to integrating the integer part function, which is what this question is about; the new title is misleading. $\endgroup$ – JiK Jul 27 '15 at 11:57
7
$\begingroup$

Notice, let $n$ be an integer then using property of definite integral, we have $$\int_{0}^{n}[x]dx=\int_{0}^{1}[x]dx+\int_{1}^{2}[x]dx+\int_{2}^{3}[x]dx+\ldots+\int_{n-1}^{n}[x]dx$$ $$=\int_{0}^{1}(0)dx+\int_{1}^{2}1dx+\int_{2}^{3}2dx+\ldots+\int_{n-1}^{n}(n-1)dx$$ $$=0+1+2+3+\ldots+(n-1)$$ $$\implies \int_{0}^{n}[x]dx=1+2+3+\ldots +(n-1)=\frac{n(n-1)}{2}$$

$\endgroup$
7
$\begingroup$

Hint: $\lfloor x \rfloor = k$ for $x \in [k,k+1)$ for $k = 0,1,2,\ldots,n-1$. Hence $$\int_0^n \lfloor x \rfloor \; dx = \sum_{k=0}^{n-1} \int_{k}^{k+1} \lfloor x \rfloor \; dx = \sum_{k=0}^{n-1} \int_{k}^{k+1} k \; dx,$$ where the integrand is a constant with respect to $x$.

$\endgroup$
-1
$\begingroup$

For any positive integer n, $[x]=n$ for $n\le x< n+1$ $$\int_0^n[x]dx=0+1+2+3+4+\ldots +(n-1)=\frac{n(n-1)}{2}$$

$\endgroup$

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.