## Jonask Group Title MVT for integrals one year ago one year ago

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 Group Title

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 Group Title

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 Group Title

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

7. amistre64 Group Title

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

okay did you have a look at the MVTS attached

9. amistre64 Group Title

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 Group Title

$f(E)=\frac{1}{b-a}\int_{a}^{b}f(x)dx$ 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 Group Title

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 Group Title

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

20. satellite73 Group Title

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

obviusly we know the primitive of cos is sin

yes

23. electrokid Group Title

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

24. electrokid Group Title

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

25. satellite73 Group Title

oh good idea!

are you talking about the equation or the last conlution

27. electrokid Group Title

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

really good idea

29. satellite73 Group Title

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

30. electrokid Group Title

yep

31. satellite73 Group Title

but unfortunately now the interval changed...

32. electrokid Group Title

so what? I = I

33. satellite73 Group Title

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

34. electrokid Group Title

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

35. satellite73 Group Title

then you don't have sine anymore

36. electrokid Group Title

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 Group Title

40. satellite73 Group Title

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 Group Title

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 Group Title

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

43. electrokid Group Title

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