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.

- anonymous

- chestercat

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- anonymous

- triciaal

sorry don't know the theorem

- anonymous

thanks for stopping by either way

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## More answers

- anonymous

@ganeshie8 do you know the theorem ?

- welshfella

|dw:1439409099284:dw|

- 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

i think thats the mean value theorem of derivatives not integrals

- welshfella

so as sqrtx is differentiable in the range [0,4] then the theorem applies

- 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

oh right sorry I misread the question I didnt see 'integrals'

- anonymous

do you know the integral one ?

- welshfella

i cant recall it i'm afraid

- anonymous

nooouuuuuu

- anonymous

@IrishBoy123 any thoughts?

- IrishBoy123

it's this
\(\displaystyle {\frac{F(b) - F(a)}{b-a}} = F '(c) [= f(c)]\)

- anonymous

does \[F(a)= \int\limits f(a)\]

- IrishBoy123

more strictly, \( \int_{a}^{b} f(x) \ dx = F(x)|_{a}^{b} = F(b) - F(a) \)

- anonymous

-16/3 ?

- IrishBoy123

|dw:1439411533571:dw|
basically you're looking for this......same areas in rectangle as function

- 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

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

okay im lost, do i find the indefinite integral of the function and plug in a and b and subtract them ?

- 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

yeah i think i got it, thank you for helping me out

- IrishBoy123

@watchpencilpaper
really sorry i was so flakey/ otherwise engaged; and hope you got something from this

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