## Jonask 2 years ago MVT for integrals

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

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. satellite73

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

7. amistre64

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

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

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

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

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

i mean def 2

15. satellite73

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

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

exacly

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

19. satellite73

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

20. satellite73

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

obviusly we know the primitive of cos is sin

yes

23. electrokid

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

24. electrokid

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

25. satellite73

oh good idea!

are you talking about the equation or the last conlution

27. electrokid

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

really good idea

29. satellite73

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

30. electrokid

yep

31. satellite73

but unfortunately now the interval changed...

32. electrokid

so what? I = I

33. satellite73

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

34. electrokid

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

35. satellite73

then you don't have sine anymore

36. electrokid

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

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

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

39. electrokid

zeta-pi/2 is your new zea

40. satellite73

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. electrokid

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

42. electrokid

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

43. electrokid

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