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FoolForMath
Albeit elementary, yet another cute integral, \[\text{ Evaluate: } \large \int \limits_0 ^{\frac 32} \lfloor x^2 \rfloor \; dx\]
This can't be continuous?
what is that box thing around x ??
round down to nearest integer or something like that?
for 0<x<1 f(x) = 0 for 1<x< 1.5 f(x) = 1 so its just 1/2 ?
what is the floor function?
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 $$(2^\frac{1}{2}-1) +2(1.5-2^\frac{1}{2})$$ = $$2-2^\frac{1}{2}$$
integrate over the sets \[[0,1],[1,\sqrt{2}],[\sqrt{2},3/2]\]
\[\int_0^1 \lfloor x^2\rfloor dx + \int_1^{\frac32}\lfloor x^2\rfloor dx\]Can I do this?
Oh there is another interval :/
oh ha now i get it
I wish wolfram showed me steps .... LOL http://www.wolframalpha.com/input/?i=inegrate+floor%28x%5E2%29+from+0+to+3%2F2
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
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.
Isn't that just \(\int\limits_{0}^{1} 0dx+ \int\limits_{1}^{\sqrt{2}}dx+\int\limits_{\sqrt{2}}^{\frac{3}{2}}2dx=(\sqrt{2}-1)+2(\frac{3}{2}-\sqrt{2})=2-\sqrt{2}\)?
Since \(⌊x^2⌋=0 \text{ for } x \in (0,1), ⌊x^2⌋=1 \text{ for } x\in (1,\sqrt{2}) \text{ and }⌊x^2⌋=2 \text{ for } x\in (\sqrt{2},\frac{3}{2}). \)
Oh Stom already did it!