Mathematics
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OpenStudy (kayders1997):
Please help volume between two curves: y=1/4x^2 and y=5-x^2
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OpenStudy (kayders1997):
@zepdrix @jim_thompson5910 @satellite73
OpenStudy (kayders1997):
@FortyTheRapper
OpenStudy (kayders1997):
@vishweshshrimali5
OpenStudy (fortytherapper):
I rather use my computer than mobile phone. Let me check if it's being used
OpenStudy (kayders1997):
Okay
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OpenStudy (kayders1997):
First I drew a sketch
OpenStudy (fortytherapper):
It is. Let me boot it up and I'll be here
OpenStudy (kayders1997):
I drew a sketch than gave up 😂
OpenStudy (anonymous):
is it \[y=\frac{1}{4}x^2\] and \[y=5-x^2\]?
OpenStudy (kayders1997):
Yeah
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OpenStudy (anonymous):
do you know where they meet up? you can kinda do it in your head
OpenStudy (kayders1997):
2
OpenStudy (kayders1997):
So x=4
OpenStudy (anonymous):
no at \(x=-2\) and at \(x=2\)
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OpenStudy (anonymous):
if that is not clear, solve \[\frac{1}{4}x^2=5-x^2\] for \(x\) put both \(-2\) and \(2\) get you \(1\)
OpenStudy (kayders1997):
Oh my bad was looking at wrong problem
OpenStudy (kayders1997):
But yes that makes sense
OpenStudy (anonymous):
since evidently \(5-x^2\) is larger, integrate \[\int_{-2}^2(5-x^2-\frac{1}{4}x^2)dx\]
OpenStudy (kayders1997):
Okay
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OpenStudy (fortytherapper):
The legend explained it perfectly
OpenStudy (kayders1997):
Lol sorry internet problems
OpenStudy (anonymous):
lol
oh just one more thing
this is area, not volume
OpenStudy (kayders1997):
Oh
OpenStudy (fortytherapper):
How's the integration?
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OpenStudy (kayders1997):
Well