Find the perimeter of the region bounded by y=x and y=x^2

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Find the perimeter of the region bounded by y=x and y=x^2

Mathematics
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Hey u r pink :P
yeah, the downside of being level 80 lol do you know the arc length formula?
yes

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Other answers:

|dw:1328027797447:dw|
so the problem is integrating it?
umm no i need help setting up the equation
umm just a lil confused
|dw:1328027896437:dw|here's you pic, intersection (1,1) the length of the y=x can be found with the pythagorean theroem...
sqrt(2) :)
ya that is what i got for the function y=x
the other one is a lil more scary
\[\int^{1}_{0} \sqrt{1+(2x)^2}dx\]
\[\frac{dy}{dx}=2x\]\[L=\int\sqrt{1+(\frac{dy}{dx})^2}dx=\int_{0}^{1}\sqrt{1+4x^2}dx\]off the bat I'm thinking trig sub\[x=\frac12\tan\theta\to dx=\frac12\sec^2\theta d\theta\]
yep, lengths go to trig stuff usually
so you wind up with\[\int_{0}^{1}\sec^3\theta d\theta\]which is a pretty well-known integral
ok but what abt the second function?
ive never meet that integral in me life :)
the second function is just the distance from 0,0 to 1,1
lol
Yesterday someone told me you need to find the arc lengths of both functions
if need be: \[\int_{0}^{1}\sqrt{1+(1)^2}dx\]
it's a little tricky the first time you see it. It can be done by integration by parts. You need to know how to do the integral of secant though
ummm i guess i can just solve it numerically if need be
talking about integral sec cubed here example 10 http://tutorial.math.lamar.edu/Classes/CalcII/IntegralsWithTrig.aspx
takes too long to type myself
\[ \int \sec x \;dx = \ln |\sec x +\tan x| +c \] I remember correctly?
you do :D
for a sec yes; tricky mathikers i tell ya
Great :D
ummm so that wld be the answer? like i forget abt the other function?
Well you have the straight part... I don't really have time to go through the whole undoing the trig sub thing, I gotta get to class
LOL ok Thanks
but one last question though
like y is that the perimeter?
we only found the arc length of the line but not the curve
|dw:1328028665611:dw|here's the triangle for your trig sub I set up the curved part for you, remember?\[\frac12\tan\theta=\ x\]\[\frac12\sec^2\theta d\theta=dx\]winds up as \[\int\sec^3\theta d\theta\]Use example 10http://tutorial.math.lamar.edu/Classes/CalcII/IntegralsWithTrig.aspx and add that to the straight part
oh my triangle is wrong :/
|dw:1328028961912:dw|there we go
Btw Turing Pippa is right, this integral gives the length of the parabola \(y=x^2\) right?
...bounds 0 to 1
We still need to sum up the straight line right?
I think amistre did that two different ways above
ohhh ok i see
yes that is correct LOL
I forgot abt that :D
pythagorean theorem, then he used arc length same answer for the straight bit both ways
Alrighty Thanks :D Dont be late for class
yeah, 2 min late already!
I don't understand, :(
later fellas! @FFM figure it out :P
hehe thanks
Duh got it :D It's simple Pythagorean theorem no need of integrating ;)
hehehe I dont follow that
ya but i get what to do
peepppaaa :D
lol yes

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