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