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Jonask
 one year ago
Best ResponseYou've already chosen the best response.0prove 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
 one year ago
Best ResponseYou've already chosen the best response.0integrate the left side to determine its value

Jonask
 one year ago
Best ResponseYou've already chosen the best response.0i used by parts and i have \[2I=e^{x}(\sin x\cos x)_0^\pi\]

amistre64
 one year ago
Best ResponseYou've already chosen the best response.0e^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

satellite73
 one year ago
Best ResponseYou've already chosen the best response.0my guess is that you do not have to compute anything (but i could be wrong)

amistre64
 one year ago
Best ResponseYou've already chosen the best response.0..... but .... i like computing :)

Jonask
 one year ago
Best ResponseYou've already chosen the best response.0okay did you have a look at the MVTS attached

amistre64
 one year ago
Best ResponseYou've already chosen the best response.0cosx changes sign in the interval, e^x stays positive tho so we can let g(x)=e^x

Jonask
 one year ago
Best ResponseYou've already chosen the best response.0plugging in the boundaries gives \[\frac{1}{2}(e^{\pi}+1)\le 2(1/2)\le 1\]

Jonask
 one year ago
Best ResponseYou've already chosen the best response.0i agree with @satellite73 in mean value teorems theres very little comp.

amistre64
 one year ago
Best ResponseYou've already chosen the best response.0\[f(E)=\frac{1}{ba}\int_{a}^{b}f(x)dx\] might be helpful? or i mightbe barking up the wrong tree

Jonask
 one year ago
Best ResponseYou've already chosen the best response.0well since we have two functions i figured that we might need a defination with f and g

satellite73
 one year ago
Best ResponseYou've already chosen the best response.0you 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
 one year ago
Best ResponseYou've already chosen the best response.0ane other helpful toll \[m(ba) \le \int\limits _a^b f(x)dx \le M(ba)\]

Jonask
 one year ago
Best ResponseYou've already chosen the best response.0i assume C is\[e^{ \xi}\]

satellite73
 one year ago
Best ResponseYou've already chosen the best response.0\(C=\int_0^{\pi}=e^{x}dx\)

satellite73
 one year ago
Best ResponseYou've already chosen the best response.0i mean \[C=\int_0^{\pi}e^{x}dx\]

Jonask
 one year ago
Best ResponseYou've already chosen the best response.0obviusly we know the primitive of cos is sin

electrokid
 one year ago
Best ResponseYou've already chosen the best response.1how about replacing x with x=upi/2?

electrokid
 one year ago
Best ResponseYou've already chosen the best response.1then cos becomes sin, exp(x) can still be taken as g(x)...

Jonask
 one year ago
Best ResponseYou've already chosen the best response.0are you talking about the equation or the last conlution

electrokid
 one year ago
Best ResponseYou've already chosen the best response.1in the original eq.. then follow the same steps as above...

satellite73
 one year ago
Best ResponseYou've already chosen the best response.0\[\int_{\frac{\pi}{2}}^{\frac{\pi}{2}}e^{x\frac{\pi}{2}}\sin(x)dx\]

satellite73
 one year ago
Best ResponseYou've already chosen the best response.0but unfortunately now the interval changed...

satellite73
 one year ago
Best ResponseYou've already chosen the best response.0yeah but how do you know \[0<\xi<\pi\]??

electrokid
 one year ago
Best ResponseYou've already chosen the best response.1then, zeta lies in [pi/2,pi/2] then translate the coordinates back to "x"

satellite73
 one year ago
Best ResponseYou've already chosen the best response.0then you don't have sine anymore

electrokid
 one year ago
Best ResponseYou've already chosen the best response.1you are looking at the same space but you have repositioned yourself at a different point is space...

Jonask
 one year ago
Best ResponseYou've already chosen the best response.0\[0\le \xi\pi/2\le \pi\]

Jonask
 one year ago
Best ResponseYou've already chosen the best response.0\[0\le \xi+\pi/2\le \pi\]

electrokid
 one year ago
Best ResponseYou've already chosen the best response.1zetapi/2 is your new zea

satellite73
 one year ago
Best ResponseYou've already chosen the best response.0maybe 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
 one year ago
Best ResponseYou've already chosen the best response.1fine name the one obtained in "UV" plane as "eta" then that same "eta" is "zeta" in "XY" plane... function did not change

electrokid
 one year ago
Best ResponseYou've already chosen the best response.1\[ {\pi\over2}\le\eta\le{\pi\over2},\qquad{\pi\over2}\le u\le{\pi\over2}\\ 0\le\eta+{\pi\over2}\le\pi,\qquad0\le u+{\pi\over2}\le\pi\\ 0\le\zeta\pi,\qquad0\le x\le\pi \] we had

electrokid
 one year ago
Best ResponseYou've already chosen the best response.1a 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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