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

1. Ishaan94 Group Title

This can't be continuous?

2. FoolForMath Group Title

why can't it be?

3. Ishaan94 Group Title

idk hehe sorry.

4. experimentX Group Title

what is that box thing around x ??

5. Zarkon Group Title

floor function

6. eigenschmeigen Group Title

round down to nearest integer or something like that?

7. eigenschmeigen Group Title

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

8. Zarkon Group Title

no

9. eigenschmeigen Group Title

what is the floor function?

10. Zarkon Group Title
11. Stom Group Title

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 Group Title

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

13. Zarkon Group Title

that is correct

14. Ishaan94 Group Title

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

15. Ishaan94 Group Title

Oh there is another interval :/

16. experimentX Group Title
17. eigenschmeigen Group Title

oh ha now i get it

18. experimentX Group Title

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 Group Title

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 Group Title

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 Group Title

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 Group Title

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 Group Title