A community for students.
Here's the question you clicked on:
 0 viewing
FoolForMath
 3 years ago
Albeit elementary, yet another cute integral,
\[\text{ Evaluate: } \large \int \limits_0 ^{\frac 32} \lfloor x^2 \rfloor \; dx\]
FoolForMath
 3 years ago
Albeit elementary, yet another cute integral, \[\text{ Evaluate: } \large \int \limits_0 ^{\frac 32} \lfloor x^2 \rfloor \; dx\]

This Question is Closed

Ishaan94
 3 years ago
Best ResponseYou've already chosen the best response.0This can't be continuous?

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

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

eigenschmeigen
 3 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 ?

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

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

Ishaan94
 3 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?

Ishaan94
 3 years ago
Best ResponseYou've already chosen the best response.0Oh there is another interval :/

eigenschmeigen
 3 years ago
Best ResponseYou've already chosen the best response.0oh ha now i get it

experimentX
 3 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

Ishaan94
 3 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
 3 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
 3 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
 3 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
 3 years ago
Best ResponseYou've already chosen the best response.0Oh Stom already did it!
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.