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Calculus1 22 Online
OpenStudy (anonymous):

let f be a continious function on the closed interval [0,2]. If 2<(or equal to)f(x)<(or equal to)4, then the greatest possible value of (integral sign from 0 to 2) f(x)dx is?

OpenStudy (anonymous):

the least it could be is \[\int_0^22dx=2\times 2=4\]

OpenStudy (anonymous):

how did you get that?

OpenStudy (loser66):

I don't know either hehehe

OpenStudy (anonymous):

how did i get the integral?

OpenStudy (anonymous):

more like where did the 2dx come from?

OpenStudy (anonymous):

the least the function can be is 2, so i imagined it was always 2, i.e. it was \(f(x)=2\)

OpenStudy (anonymous):

then the integral is base times height, i get \(2\times 2=4\) for the minimum possible value of the integral if the function was always \(f(x)=4\) then \(2\times 4=8\) is the maximum

OpenStudy (anonymous):

why is there a base times height thing?

OpenStudy (loser66):

oh I got it now.

OpenStudy (loser66):

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OpenStudy (loser66):

|dw:1398830286951:dw|

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