Question

question

Answer 1

that is the greatest integer function btw..

Answer 2

how do i start?

Answer 3

Consider splitting up the integral.

Answer 4

how do i split it up?

Answer 5

\[
\int_0^k \lfloor x \rfloor\;dx=\int_0^1 \lfloor x \rfloor\;dx+\int_1^k \lfloor x \rfloor\;dx = \int_0^1x\;dx+\int_1^k \lfloor x \rfloor\;dx
\]

Answer 6

Doing it a second time \[
\int_1^2\lfloor x\rfloor \;dx = \int_1^2 x- 1 \;dx = \int_0^1 x \;dx
\]

Answer 7

If you do it \(k\) times you end up with: \[
\int_0^k \lfloor x\rfloor \;dx= k\int_0^1x\;dx
\]

Answer 8

how did you come up with the \(2\)? replace it with the \(k\)?

Answer 9

No, I split up the integral like so:\[
\int_0^k=\underbrace{\int_0^1+\int_1^2+\int_2^3+\ldots + \int_{k-2}^{k-1}+\int_{k-1}^k}_{k}
\]

Answer 10

okay, i got it, how did you get the \(x-1\) when you did it the second time?

Answer 11

Then I look at any arbitrary integral in there: \[
\int_a^{a+1}\lfloor x \rfloor \;dx = \int_a^{a+1}a\;dx
\]

Answer 12

I suppose each term actually becomes: \[
ax\Bigg|_a^{a+1}
\]

Answer 13

@mathsnerd101 I suppose I was a bit wrong earlier.

Answer 14

the first step was wrong?

Answer 15

Evaluating any of those integrals gives you \[
\int_a^{a+1}\lfloor x\rfloor\;dx=a(a+1)-a(a) = a
\]

Answer 16

but it's \(k\) on top..

Answer 17

So you really just end up with: \[
\int_0^k\lfloor x\rfloor \;dx = \sum_{a=0}^k a
\]

Answer 18

@mathsnerd101 I know, but you want to integrate between each single integer.

Answer 19

The reason why is because the greatest integer function is only continuous between two integers, and you can only integrate on continuous intervals.

Answer 20

so you cant integrate it?

Answer 21

No, you can integrate it, but you have to split up the integral like I did.

Answer 22

and integrate it one by one?

Answer 23

You can integrate between 4 and 5, but not between 3 and 5

Answer 24

Yes, that's why I integrate by a and a+1

Answer 25

why not 3 and 5?

Answer 26

Because at 3, 4, and 5, it is discontinuous. You cant have that discontinuity of 4 right in the middle

Answer 27

ok, i got it, thank you!