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Jonask

  • 2 years ago

MVT for integrals

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

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

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

  5. amistre64
    • 2 years ago
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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

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

  7. amistre64
    • 2 years ago
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    ..... but .... i like computing :)

  8. Jonask
    • 2 years ago
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    okay did you have a look at the MVTS attached

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

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

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

  12. amistre64
    • 2 years ago
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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

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

  14. Jonask
    • 2 years ago
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    i mean def 2

  15. satellite73
    • 2 years ago
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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

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

  17. Jonask
    • 2 years ago
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    exacly

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

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

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

  21. Jonask
    • 2 years ago
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    obviusly we know the primitive of cos is sin

  22. Jonask
    • 2 years ago
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    yes

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

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

  25. satellite73
    • 2 years ago
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    oh good idea!

  26. Jonask
    • 2 years ago
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    are you talking about the equation or the last conlution

  27. electrokid
    • 2 years ago
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    in the original eq.. then follow the same steps as above...

  28. Jonask
    • 2 years ago
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    really good idea

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

  30. electrokid
    • 2 years ago
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    yep

  31. satellite73
    • 2 years ago
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    but unfortunately now the interval changed...

  32. electrokid
    • 2 years ago
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    so what? I = I

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

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

  35. satellite73
    • 2 years ago
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    then you don't have sine anymore

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

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

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

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

  40. satellite73
    • 2 years ago
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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

  41. electrokid
    • 2 years ago
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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

  42. electrokid
    • 2 years ago
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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

  43. electrokid
    • 2 years ago
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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)

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