Question

Albeit elementary, yet another cute integral,

\[\text{ Evaluate: } \large \int \limits_0 ^{\frac 32} \lfloor x^2 \rfloor \; dx\]

Answer 1

This can't be continuous?

Answer 2

why can't it be?

Answer 3

idk hehe sorry.

Answer 4

what is that box thing around x ??

Answer 5

floor function

Answer 6

round down to nearest integer or something like that?

Answer 7

for 0<x<1 f(x) = 0

for 1<x< 1.5 f(x) = 1

so its just 1/2 ?

Answer 8

no

Answer 9

what is the floor function?

Answer 10

http://en.wikipedia.org/wiki/Floor_and_ceiling_functions

Answer 11

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
$$(2^\frac{1}{2}-1) +2(1.5-2^\frac{1}{2})$$
=
$$2-2^\frac{1}{2}$$

Answer 12

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

Answer 13

that is correct

Answer 14

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

Answer 15

Oh there is another interval :/

Answer 16

http://www.wolframalpha.com/input/?i=plot+floor%28x%5E2%29

Answer 17

oh ha now i get it

Answer 18

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

Answer 19

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

Answer 20

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.

Answer 21

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

Answer 22

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

Answer 23

Oh Stom already did it!