## anonymous 3 years ago MVT for integrals

1. anonymous

prove that there exist a $\xi$ in $0 \le x \le \pi$ such that $\huge \int\limits _0^\pi e^{-x}\cos x dx =\sin \xi$

2. amistre64

integrate the left side to determine its value

3. anonymous

4. anonymous

i used by parts and i have $2I=e^{-x}(\sin x-\cos x)|_0^\pi$

5. amistre64

e^-x +cosx -e^-x - -sinx e^-x + -cosx -e^-x $\int e^{-x}\cos x dx =-cos(x)~e^{-x}+sin(x)e^{-x}...$that looks about right so far

6. anonymous

my guess is that you do not have to compute anything (but i could be wrong)

7. amistre64

..... but .... i like computing :)

8. anonymous

okay did you have a look at the MVTS attached

9. amistre64

cosx changes sign in the interval, e^-x stays positive tho so we can let g(x)=e^-x

10. anonymous

plugging in the boundaries gives $\frac{1}{2}(e^{-\pi}+1)\le 2(1/2)\le 1$

11. anonymous

i agree with @satellite73 in mean value teorems theres very little comp.

12. amistre64

$f(E)=\frac{1}{b-a}\int_{a}^{b}f(x)dx$ might be helpful? or i mightbe barking up the wrong tree

13. anonymous

well since we have two functions i figured that we might need a defination with f and g

14. anonymous

i mean def 2

15. anonymous

you get directly that $\int_0^{\pi}e^{-x}\cos(x)dx=\cos(\xi)C$ where $$C<1$$ but i don't see how to get a sine out of it

16. anonymous

ane other helpful toll $m(b-a) \le \int\limits _a^b f(x)dx \le M(b-a)$

17. anonymous

exacly

18. anonymous

i assume C is$e^{- \xi}$

19. anonymous

$$C=\int_0^{\pi}=e^{-x}dx$$

20. anonymous

i mean $C=\int_0^{\pi}e^{-x}dx$

21. anonymous

obviusly we know the primitive of cos is sin

22. anonymous

yes

23. anonymous

how about replacing x with x=u-pi/2?

24. anonymous

then cos becomes sin, exp(-x) can still be taken as g(x)...

25. anonymous

oh good idea!

26. anonymous

are you talking about the equation or the last conlution

27. anonymous

in the original eq.. then follow the same steps as above...

28. anonymous

really good idea

29. anonymous

$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}e^{x-\frac{\pi}{2}}\sin(x)dx$

30. anonymous

yep

31. anonymous

but unfortunately now the interval changed...

32. anonymous

so what? I = I

33. anonymous

yeah but how do you know $0<\xi<\pi$??

34. anonymous

then, zeta lies in [-pi/2,pi/2] then translate the co-ordinates back to "x"

35. anonymous

then you don't have sine anymore

36. anonymous

you are looking at the same space but you have repositioned yourself at a different point is space...

37. anonymous

$0\le \xi-\pi/2\le \pi$

38. anonymous

$0\le \xi+\pi/2\le \pi$

39. anonymous

40. anonymous

maybe i am confused, but if you change the interval to get sine, i don't see how you can change back and still keep sine the theorem says there is a $$\xi$$ in $$(a,b)$$ and you changed the interval to turn cosine in to sine

41. anonymous

fine name the one obtained in "UV" plane as "eta" then that same "eta" is "zeta" in "XY" plane... function did not change

42. anonymous

$-{\pi\over2}\le\eta\le{\pi\over2},\qquad-{\pi\over2}\le u\le{\pi\over2}\\ 0\le\eta+{\pi\over2}\le\pi,\qquad0\le u+{\pi\over2}\le\pi\\ 0\le\zeta\pi,\qquad0\le x\le\pi$ we had

43. anonymous

a great man once said... math becomes very meaningful when you keep the mathematical meanings out!!! it will lead to logic :) (me)