anonymous
  • anonymous
MVT for integrals
Mathematics
  • Stacey Warren - Expert brainly.com
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
prove that there exist a \[\xi\] in \[0 \le x \le \pi\] such that \[\huge \int\limits _0^\pi e^{-x}\cos x dx =\sin \xi \]
amistre64
  • amistre64
integrate the left side to determine its value
anonymous
  • anonymous
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anonymous
  • anonymous
i used by parts and i have \[2I=e^{-x}(\sin x-\cos x)|_0^\pi\]
amistre64
  • 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
anonymous
  • anonymous
my guess is that you do not have to compute anything (but i could be wrong)
amistre64
  • amistre64
..... but .... i like computing :)
anonymous
  • anonymous
okay did you have a look at the MVTS attached
amistre64
  • amistre64
cosx changes sign in the interval, e^-x stays positive tho so we can let g(x)=e^-x
anonymous
  • anonymous
plugging in the boundaries gives \[\frac{1}{2}(e^{-\pi}+1)\le 2(1/2)\le 1\]
anonymous
  • anonymous
i agree with @satellite73 in mean value teorems theres very little comp.
amistre64
  • amistre64
\[f(E)=\frac{1}{b-a}\int_{a}^{b}f(x)dx\] might be helpful? or i mightbe barking up the wrong tree
anonymous
  • anonymous
well since we have two functions i figured that we might need a defination with f and g
anonymous
  • anonymous
i mean def 2
anonymous
  • 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
anonymous
  • anonymous
ane other helpful toll \[m(b-a) \le \int\limits _a^b f(x)dx \le M(b-a)\]
anonymous
  • anonymous
exacly
anonymous
  • anonymous
i assume C is\[e^{- \xi}\]
anonymous
  • anonymous
\(C=\int_0^{\pi}=e^{-x}dx\)
anonymous
  • anonymous
i mean \[C=\int_0^{\pi}e^{-x}dx\]
anonymous
  • anonymous
obviusly we know the primitive of cos is sin
anonymous
  • anonymous
yes
anonymous
  • anonymous
how about replacing x with x=u-pi/2?
anonymous
  • anonymous
then cos becomes sin, exp(-x) can still be taken as g(x)...
anonymous
  • anonymous
oh good idea!
anonymous
  • anonymous
are you talking about the equation or the last conlution
anonymous
  • anonymous
in the original eq.. then follow the same steps as above...
anonymous
  • anonymous
really good idea
anonymous
  • anonymous
\[\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}e^{x-\frac{\pi}{2}}\sin(x)dx\]
anonymous
  • anonymous
yep
anonymous
  • anonymous
but unfortunately now the interval changed...
anonymous
  • anonymous
so what? I = I
anonymous
  • anonymous
yeah but how do you know \[0<\xi<\pi\]??
anonymous
  • anonymous
then, zeta lies in [-pi/2,pi/2] then translate the co-ordinates back to "x"
anonymous
  • anonymous
then you don't have sine anymore
anonymous
  • anonymous
you are looking at the same space but you have repositioned yourself at a different point is space...
anonymous
  • anonymous
\[0\le \xi-\pi/2\le \pi\]
anonymous
  • anonymous
\[0\le \xi+\pi/2\le \pi\]
anonymous
  • anonymous
zeta-pi/2 is your new zea
anonymous
  • 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
anonymous
  • anonymous
fine name the one obtained in "UV" plane as "eta" then that same "eta" is "zeta" in "XY" plane... function did not change
anonymous
  • 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
anonymous
  • anonymous
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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