## FoolForMath 3 years ago Albeit elementary, yet another cute integral, $\text{ Evaluate: } \large \int \limits_0 ^{\frac 32} \lfloor x^2 \rfloor \; dx$

1. Ishaan94

This can't be continuous?

2. FoolForMath

why can't it be?

3. Ishaan94

idk hehe sorry.

4. experimentX

what is that box thing around x ??

5. Zarkon

floor function

6. eigenschmeigen

round down to nearest integer or something like that?

7. eigenschmeigen

for 0<x<1 f(x) = 0 for 1<x< 1.5 f(x) = 1 so its just 1/2 ?

8. Zarkon

no

9. eigenschmeigen

what is the floor function?

10. Zarkon
11. Stom

can we do it like this,breaking the integral in 3 parts from 0 to 1 ------>> the value inside integral is 0 from 1 to root (2) ------>> the value inside integral is 1 and from root2 to 1.5 ------>> the value inside integral is 2 so the answer is $$(2^\frac{1}{2}-1) +2(1.5-2^\frac{1}{2})$$ = $$2-2^\frac{1}{2}$$

12. Zarkon

integrate over the sets $[0,1],[1,\sqrt{2}],[\sqrt{2},3/2]$

13. Zarkon

that is correct

14. Ishaan94

$\int_0^1 \lfloor x^2\rfloor dx + \int_1^{\frac32}\lfloor x^2\rfloor dx$Can I do this?

15. Ishaan94

Oh there is another interval :/

16. experimentX
17. eigenschmeigen

oh ha now i get it

18. experimentX

I wish wolfram showed me steps .... LOL http://www.wolframalpha.com/input/?i=inegrate+floor%28x%5E2%29+from+0+to+3%2F2

19. Ishaan94

No, we don't need electronic aid. I should have applied my head root2 is around 1.4 I think, less than 1.5. Or a better way is to check at which points the function gets to 2. If its less than 1.5 you split it. maybe

20. KingGeorge

Using Stom's work, I'm getting $(2^{1/2}-1)+2({3/2}-2^{1/2})=2^{1/2}-1+3-2^{3/2}$So we have $2+(2^{1/2}-2^{3/2})=2-2^{1/2}=2-\sqrt2$ Which is just what Stom had.

21. Mr.Math

Isn't that just $$\int\limits_{0}^{1} 0dx+ \int\limits_{1}^{\sqrt{2}}dx+\int\limits_{\sqrt{2}}^{\frac{3}{2}}2dx=(\sqrt{2}-1)+2(\frac{3}{2}-\sqrt{2})=2-\sqrt{2}$$?

22. Mr.Math

Since $$⌊x^2⌋=0 \text{ for } x \in (0,1), ⌊x^2⌋=1 \text{ for } x\in (1,\sqrt{2}) \text{ and }⌊x^2⌋=2 \text{ for } x\in (\sqrt{2},\frac{3}{2}).$$

23. Mr.Math