anonymous
  • anonymous
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 x-coordinates of the point(s) guaranteed to exist by the theorem.
Mathematics
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chestercat
  • chestercat
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anonymous
  • anonymous
@IrishBoy123 @triciaal @Hero @welshfella
triciaal
  • triciaal
sorry don't know the theorem
anonymous
  • anonymous
thanks for stopping by either way

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anonymous
  • anonymous
@ganeshie8 do you know the theorem ?
welshfella
  • welshfella
|dw:1439409099284:dw|
welshfella
  • welshfella
basically 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
  • anonymous
i think thats the mean value theorem of derivatives not integrals
welshfella
  • welshfella
so as sqrtx is differentiable in the range [0,4] then the theorem applies
IrishBoy123
  • IrishBoy123
hi @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
  • welshfella
oh right sorry I misread the question I didnt see 'integrals'
anonymous
  • anonymous
do you know the integral one ?
welshfella
  • welshfella
i cant recall it i'm afraid
anonymous
  • anonymous
nooouuuuuu
anonymous
  • anonymous
@IrishBoy123 any thoughts?
IrishBoy123
  • IrishBoy123
it's this \(\displaystyle {\frac{F(b) - F(a)}{b-a}} = F '(c) [= f(c)]\)
anonymous
  • anonymous
does \[F(a)= \int\limits f(a)\]
IrishBoy123
  • IrishBoy123
more strictly, \( \int_{a}^{b} f(x) \ dx = F(x)|_{a}^{b} = F(b) - F(a) \)
anonymous
  • anonymous
-16/3 ?
IrishBoy123
  • IrishBoy123
|dw:1439411533571:dw| basically you're looking for this......same areas in rectangle as function
IrishBoy123
  • IrishBoy123
but 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
  • anonymous
umm 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
  • anonymous
okay im lost, do i find the indefinite integral of the function and plug in a and b and subtract them ?
IrishBoy123
  • IrishBoy123
in 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
  • anonymous
yeah i think i got it, thank you for helping me out
IrishBoy123
  • IrishBoy123
@watchpencilpaper really sorry i was so flakey/ otherwise engaged; and hope you got something from this

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