Definite Integral

  • #1
1,789
4
Hi

I need help doing the following integration:

[tex]\int_{x=0}^{x=n}[{x-\frac{1}{\sqrt{2}}]-[{x-\frac{1}{\sqrt{3}}]dx[/tex]

where n is an integer and [.] denotes the greatest integer function (floor), i.e. [x] = greatest integer less than or equal to x.

The answer given in the book is

[tex]n(\frac{1}{\sqrt{3}}-\frac{1}{\sqrt{2}})[/tex]

whereas I am getting zero as the answer. My solution is a bit jerky due to the step marked # below:

By definition, there exists some [tex]k \epsilon Z[/tex] such that

[tex]k\leq x < k+1[/tex] (so that [x] = k)

which means that

[tex]k-1<k-\frac{1}{\sqrt{2}}\leq x - \frac{1}{\sqrt{2}} < k+1-\frac{1}{\sqrt{2}}[/tex] and
[tex]k-1<k-\frac{1}{\sqrt{3}}\leq x - \frac{1}{\sqrt{3}} < k+1-\frac{1}{\sqrt{3}}[/tex]

But this would mean that (#)

[tex][{x-\frac{1}{\sqrt{2}}] = k-1[/tex]
[tex][{x-\frac{1}{\sqrt{3}}] = k-1[/tex]

Making the integrand zero and hence the integral zero as well.

I am not sure if this reasoning is correct (in particular, the integer parts cannot be greater than k-1 so this step could be wrong but still they can attain no other integral value) so I would be very grateful if someone could guide me here.

Thanks and cheers
Vivek
 
Last edited:

Answers and Replies

  • #2
Galileo
Science Advisor
Homework Helper
1,989
6
I suggest another method of approach. Draw the graph. Once you have that, the answer is trivial.

Notice that
[tex]\lfloor x-\sqrt2 \rfloor=\lfloor x-\sqrt3 \rfloor[/tex] if [tex]0\leq x < \sqrt{2}[/tex] or [tex]\sqrt{3}\leq x \leq 1[/tex]

What happens when [tex]\sqrt{2} \leq x < \sqrt{3}[/tex].

What changes when x increases by 1?
 
Last edited:

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