## 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.

1. anonymous

@IrishBoy123 @triciaal @Hero @welshfella

2. triciaal

sorry don't know the theorem

3. anonymous

thanks for stopping by either way

4. anonymous

@ganeshie8 do you know the theorem ?

5. welshfella

|dw:1439409099284:dw|

6. 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

7. anonymous

i think thats the mean value theorem of derivatives not integrals

8. welshfella

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

9. 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

10. welshfella

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

11. anonymous

do you know the integral one ?

12. welshfella

i cant recall it i'm afraid

13. anonymous

nooouuuuuu

14. anonymous

@IrishBoy123 any thoughts?

15. IrishBoy123

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

16. anonymous

does $F(a)= \int\limits f(a)$

17. IrishBoy123

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

18. anonymous

-16/3 ?

19. IrishBoy123

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

20. 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......

21. 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 ?

22. anonymous

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

23. 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"

24. anonymous

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

25. IrishBoy123

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