## mlddmlnog 2 years ago find all values x* described in the mean value theorem for integrals. f(x)=s/x^2 [1,3]

1. mlddmlnog

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2. mlddmlnog

@zepdrix

3. amoodarya

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4. amoodarya

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5. mlddmlnog

6. amoodarya

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7. mlddmlnog

i'still not comprehending... it's really hard to understand your work.. :"(

8. amoodarya

9. mlddmlnog

thank you.. but anyone else know how ot do this differently? or clearly? :D

10. amoodarya

which part of that is complicated for you ?

11. mlddmlnog

the formula you used, which is different from what i have on my review sheet, and i still can't read someof your work :'(

12. mlddmlnog

ah.., let me try this myself :) i think i now get what you did!

13. amoodarya

14. mlddmlnog

ok, i dont.get.this. where did the "S" come from?

15. amoodarya

s in figure is area of f(x)

16. mlddmlnog

could you try doing it without the "s"?

17. amoodarya

yeah without s i.e. f(x)= 1/x^2 ?

18. mlddmlnog

yes :)

19. amoodarya

20. amoodarya

look if ok go on

21. mlddmlnog

oh... ok. im gonna have lf again. thank you:)

22. mlddmlnog

i meant try myself again.

23. amoodarya

do you know the mean value of integrals theory?

24. mlddmlnog

ok, so here is what i did. $\frac{ 1 }{ 3-1 }\int\limits_{1}^{3}\frac{ 1 }{ x ^{x} }dx = \frac{ 1 }{ 2 }\int\limits_{1}^{3}x ^{-2}dx ^{}$ $\frac{ 1 }{ 2 }(-x ^{-1})$from 1 to 3, $\frac{ 1 }{ 2 }(-3^{-1}+1^{-1}) = \frac{ 1 }{ 2 }(\frac{ 1 }{ -3 }+\frac{ 1 }{ 1 }) = \frac{ 1 }{ 2 }(\frac{ 2 }{ 3 })=\frac{ 1 }{ 3 }$

25. mlddmlnog

and i dont know what to do next. but i do know that i have found(x*) so far.