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anonymous
 4 years ago
Albeit elementary, yet another cute integral,
\[\text{ Evaluate: } \large \int \limits_0 ^{\frac 32} \lfloor x^2 \rfloor \; dx\]
anonymous
 4 years ago
Albeit elementary, yet another cute integral, \[\text{ Evaluate: } \large \int \limits_0 ^{\frac 32} \lfloor x^2 \rfloor \; dx\]

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0This can't be continuous?

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.0what is that box thing around x ??

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0round down to nearest integer or something like that?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0for 0<x<1 f(x) = 0 for 1<x< 1.5 f(x) = 1 so its just 1/2 ?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0what is the floor function?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0can 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 $$(2^\frac{1}{2}1) +2(1.52^\frac{1}{2})$$ = $$22^\frac{1}{2}$$

Zarkon
 4 years ago
Best ResponseYou've already chosen the best response.2integrate over the sets \[[0,1],[1,\sqrt{2}],[\sqrt{2},3/2]\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\int_0^1 \lfloor x^2\rfloor dx + \int_1^{\frac32}\lfloor x^2\rfloor dx\]Can I do this?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Oh there is another interval :/

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.0I wish wolfram showed me steps .... LOL http://www.wolframalpha.com/input/?i=inegrate+floor%28x%5E2%29+from+0+to+3%2F2

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0No, 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

KingGeorge
 4 years ago
Best ResponseYou've already chosen the best response.0Using Stom's work, I'm getting \[(2^{1/2}1)+2({3/2}2^{1/2})=2^{1/2}1+32^{3/2}\]So we have \[2+(2^{1/2}2^{3/2})=22^{1/2}=2\sqrt2\] Which is just what Stom had.

Mr.Math
 4 years ago
Best ResponseYou've already chosen the best response.0Isn't that just \(\int\limits_{0}^{1} 0dx+ \int\limits_{1}^{\sqrt{2}}dx+\int\limits_{\sqrt{2}}^{\frac{3}{2}}2dx=(\sqrt{2}1)+2(\frac{3}{2}\sqrt{2})=2\sqrt{2}\)?

Mr.Math
 4 years ago
Best ResponseYou've already chosen the best response.0Since \(⌊x^2⌋=0 \text{ for } x \in (0,1), ⌊x^2⌋=1 \text{ for } x\in (1,\sqrt{2}) \text{ and }⌊x^2⌋=2 \text{ for } x\in (\sqrt{2},\frac{3}{2}). \)

Mr.Math
 4 years ago
Best ResponseYou've already chosen the best response.0Oh Stom already did it!
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