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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 x-coordinates of the point(s) guaranteed to exist by the theorem.

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  1. anonymous
    • one year ago
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    @IrishBoy123 @triciaal @Hero @welshfella

  2. triciaal
    • one year ago
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    sorry don't know the theorem

  3. anonymous
    • one year ago
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    thanks for stopping by either way

  4. anonymous
    • one year ago
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    @ganeshie8 do you know the theorem ?

  5. welshfella
    • one year ago
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    |dw:1439409099284:dw|

  6. welshfella
    • one year ago
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    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

  7. anonymous
    • one year ago
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    i think thats the mean value theorem of derivatives not integrals

  8. welshfella
    • one year ago
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    so as sqrtx is differentiable in the range [0,4] then the theorem applies

  9. IrishBoy123
    • one year ago
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    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

  10. welshfella
    • one year ago
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    oh right sorry I misread the question I didnt see 'integrals'

  11. anonymous
    • one year ago
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    do you know the integral one ?

  12. welshfella
    • one year ago
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    i cant recall it i'm afraid

  13. anonymous
    • one year ago
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    nooouuuuuu

  14. anonymous
    • one year ago
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    @IrishBoy123 any thoughts?

  15. IrishBoy123
    • one year ago
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    it's this \(\displaystyle {\frac{F(b) - F(a)}{b-a}} = F '(c) [= f(c)]\)

  16. anonymous
    • one year ago
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    does \[F(a)= \int\limits f(a)\]

  17. IrishBoy123
    • one year ago
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    more strictly, \( \int_{a}^{b} f(x) \ dx = F(x)|_{a}^{b} = F(b) - F(a) \)

  18. anonymous
    • one year ago
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    -16/3 ?

  19. IrishBoy123
    • one year ago
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    |dw:1439411533571:dw| basically you're looking for this......same areas in rectangle as function

  20. IrishBoy123
    • one year ago
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    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......

  21. anonymous
    • one year ago
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    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 ?

  22. anonymous
    • one year ago
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    okay im lost, do i find the indefinite integral of the function and plug in a and b and subtract them ?

  23. IrishBoy123
    • one year ago
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    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"

  24. anonymous
    • one year ago
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    yeah i think i got it, thank you for helping me out

  25. IrishBoy123
    • one year ago
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    @watchpencilpaper really sorry i was so flakey/ otherwise engaged; and hope you got something from this

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