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Jonask

MVT for integrals

  • one year ago
  • one year ago

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

    • one year ago
  2. amistre64
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    integrate the left side to determine its value

    • one year ago
  3. Jonask
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    • one year ago
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  4. Jonask
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    i used by parts and i have \[2I=e^{-x}(\sin x-\cos x)|_0^\pi\]

    • one year ago
  5. amistre64
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    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

    • one year ago
  6. satellite73
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    my guess is that you do not have to compute anything (but i could be wrong)

    • one year ago
  7. amistre64
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    ..... but .... i like computing :)

    • one year ago
  8. Jonask
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    okay did you have a look at the MVTS attached

    • one year ago
  9. amistre64
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    cosx changes sign in the interval, e^-x stays positive tho so we can let g(x)=e^-x

    • one year ago
  10. Jonask
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    plugging in the boundaries gives \[\frac{1}{2}(e^{-\pi}+1)\le 2(1/2)\le 1\]

    • one year ago
  11. Jonask
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    i agree with @satellite73 in mean value teorems theres very little comp.

    • one year ago
  12. amistre64
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    \[f(E)=\frac{1}{b-a}\int_{a}^{b}f(x)dx\] might be helpful? or i mightbe barking up the wrong tree

    • one year ago
  13. Jonask
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    well since we have two functions i figured that we might need a defination with f and g

    • one year ago
  14. Jonask
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    i mean def 2

    • one year ago
  15. satellite73
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    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

    • one year ago
  16. Jonask
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    ane other helpful toll \[m(b-a) \le \int\limits _a^b f(x)dx \le M(b-a)\]

    • one year ago
  17. Jonask
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    exacly

    • one year ago
  18. Jonask
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    i assume C is\[e^{- \xi}\]

    • one year ago
  19. satellite73
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    \(C=\int_0^{\pi}=e^{-x}dx\)

    • one year ago
  20. satellite73
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    i mean \[C=\int_0^{\pi}e^{-x}dx\]

    • one year ago
  21. Jonask
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    obviusly we know the primitive of cos is sin

    • one year ago
  22. Jonask
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    yes

    • one year ago
  23. electrokid
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    how about replacing x with x=u-pi/2?

    • one year ago
  24. electrokid
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    then cos becomes sin, exp(-x) can still be taken as g(x)...

    • one year ago
  25. satellite73
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    oh good idea!

    • one year ago
  26. Jonask
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    are you talking about the equation or the last conlution

    • one year ago
  27. electrokid
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    in the original eq.. then follow the same steps as above...

    • one year ago
  28. Jonask
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    really good idea

    • one year ago
  29. satellite73
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    \[\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}e^{x-\frac{\pi}{2}}\sin(x)dx\]

    • one year ago
  30. electrokid
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    yep

    • one year ago
  31. satellite73
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    but unfortunately now the interval changed...

    • one year ago
  32. electrokid
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    so what? I = I

    • one year ago
  33. satellite73
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    yeah but how do you know \[0<\xi<\pi\]??

    • one year ago
  34. electrokid
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    then, zeta lies in [-pi/2,pi/2] then translate the co-ordinates back to "x"

    • one year ago
  35. satellite73
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    then you don't have sine anymore

    • one year ago
  36. electrokid
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    you are looking at the same space but you have repositioned yourself at a different point is space...

    • one year ago
  37. Jonask
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    \[0\le \xi-\pi/2\le \pi\]

    • one year ago
  38. Jonask
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    \[0\le \xi+\pi/2\le \pi\]

    • one year ago
  39. electrokid
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    zeta-pi/2 is your new zea

    • one year ago
  40. satellite73
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    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

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

    • one year ago
  42. electrokid
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    \[ -{\pi\over2}\le\eta\le{\pi\over2},\qquad-{\pi\over2}\le u\le{\pi\over2}\\ 0\le\eta+{\pi\over2}\le\pi,\qquad0\le u+{\pi\over2}\le\pi\\ 0\le\zeta\pi,\qquad0\le x\le\pi \] we had

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

    • one year ago
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