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anonymous
 one year ago
Determine if the Mean Value Theorem for Integrals applies to the function f of x equals the square root of x on the interval [0, 4]. If so, find the xcoordinates of the point(s) guaranteed to exist by the theorem.
anonymous
 one year ago
Determine if the Mean Value Theorem for Integrals applies to the function f of x equals the square root of x on the interval [0, 4]. If so, find the xcoordinates of the point(s) guaranteed to exist by the theorem.

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@IrishBoy123 @triciaal @Hero @welshfella

triciaal
 one year ago
Best ResponseYou've already chosen the best response.0sorry don't know the theorem

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0thanks for stopping by either way

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@ganeshie8 do you know the theorem ?

welshfella
 one year ago
Best ResponseYou've already chosen the best response.0dw:1439409099284:dw

welshfella
 one year ago
Best ResponseYou've already chosen the best response.0basically the mean value theorem szys that if the function is differentiable in the interval [a,b] there is a number S between a and b such that f(b)  f(a)  = f'(s) b  a geometrically that means that the tangent is parallel to the chord OB

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i think thats the mean value theorem of derivatives not integrals

welshfella
 one year ago
Best ResponseYou've already chosen the best response.0so as sqrtx is differentiable in the range [0,4] then the theorem applies

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1hi @welshfella it's the Mean Value Theorem *for Integrals* it's really the same but you need to express it in terms of areas rather than tangents

welshfella
 one year ago
Best ResponseYou've already chosen the best response.0oh right sorry I misread the question I didnt see 'integrals'

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0do you know the integral one ?

welshfella
 one year ago
Best ResponseYou've already chosen the best response.0i cant recall it i'm afraid

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@IrishBoy123 any thoughts?

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1it's this \(\displaystyle {\frac{F(b)  F(a)}{ba}} = F '(c) [= f(c)]\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0does \[F(a)= \int\limits f(a)\]

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1more strictly, \( \int_{a}^{b} f(x) \ dx = F(x)_{a}^{b} = F(b)  F(a) \)

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1dw:1439411533571:dw basically you're looking for this......same areas in rectangle as function

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1but you're in interval 0 > 4 so your answer has to be in that interval or you have messed up that's very hand wavey but i'm up to ears right now......

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0umm okay i don't know what that means but ill assume that your very busy and have to go , if you come back can you check my answer ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay im lost, do i find the indefinite integral of the function and plug in a and b and subtract them ?

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1in shorthand \(\large \int_{0}^{4} \sqrt{x} \ dx = \frac{2}{3} x^{3/2} _{0}^{4} = \frac{16}{3}\) equating areas: \(f(c) . (4  0) = \frac{16}{3} \implies f(c) = \frac{4}{3} = \sqrt{c}\implies c = \frac{16}{9}\) has that helped at all, a bit rushed i know but the idea is all here dw:1439412974179:dw because same areas under curve and inside rectangle, you get "c"

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yeah i think i got it, thank you for helping me out

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1@watchpencilpaper really sorry i was so flakey/ otherwise engaged; and hope you got something from this
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