MVT for integrals

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MVT for integrals

Mathematics
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prove that there exist a \[\xi\] in \[0 \le x \le \pi\] such that \[\huge \int\limits _0^\pi e^{-x}\cos x dx =\sin \xi \]
integrate the left side to determine its value
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i used by parts and i have \[2I=e^{-x}(\sin x-\cos x)|_0^\pi\]
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
my guess is that you do not have to compute anything (but i could be wrong)
..... but .... i like computing :)
okay did you have a look at the MVTS attached
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.
\[f(E)=\frac{1}{b-a}\int_{a}^{b}f(x)dx\] 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
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\[e^{- \xi}\]
\(C=\int_0^{\pi}=e^{-x}dx\)
i mean \[C=\int_0^{\pi}e^{-x}dx\]
obviusly we know the primitive of cos is sin
yes
how about replacing x with x=u-pi/2?
then cos becomes sin, exp(-x) can still be taken as g(x)...
oh good idea!
are you talking about the equation or the last conlution
in the original eq.. then follow the same steps as above...
really good idea
\[\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}e^{x-\frac{\pi}{2}}\sin(x)dx\]
yep
but unfortunately now the interval changed...
so what? I = I
yeah but how do you know \[0<\xi<\pi\]??
then, zeta lies in [-pi/2,pi/2] then translate the co-ordinates back to "x"
then you don't have sine anymore
you are looking at the same space but you have repositioned yourself at a different point is space...
\[0\le \xi-\pi/2\le \pi\]
\[0\le \xi+\pi/2\le \pi\]
zeta-pi/2 is your new zea
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
fine name the one obtained in "UV" plane as "eta" then that same "eta" is "zeta" in "XY" plane... function did not change
\[ -{\pi\over2}\le\eta\le{\pi\over2},\qquad-{\pi\over2}\le u\le{\pi\over2}\\ 0\le\eta+{\pi\over2}\le\pi,\qquad0\le u+{\pi\over2}\le\pi\\ 0\le\zeta\pi,\qquad0\le x\le\pi \] we had
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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