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anonymous
 3 years ago
MVT for integrals
anonymous
 3 years ago
MVT for integrals

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anonymous
 3 years 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
 3 years ago
Best ResponseYou've already chosen the best response.0integrate the left side to determine its value

anonymous
 3 years 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
 3 years 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

anonymous
 3 years 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
 3 years ago
Best ResponseYou've already chosen the best response.0..... but .... i like computing :)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0okay did you have a look at the MVTS attached

amistre64
 3 years 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

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

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

amistre64
 3 years 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

anonymous
 3 years 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

anonymous
 3 years 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

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

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

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

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0obviusly we know the primitive of cos is sin

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0how about replacing x with x=upi/2?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0then cos becomes sin, exp(x) can still be taken as g(x)...

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0are you talking about the equation or the last conlution

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0in the original eq.. then follow the same steps as above...

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0but unfortunately now the interval changed...

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0then, zeta lies in [pi/2,pi/2] then translate the coordinates back to "x"

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0then you don't have sine anymore

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0you are looking at the same space but you have repositioned yourself at a different point is space...

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[0\le \xi\pi/2\le \pi\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[0\le \xi+\pi/2\le \pi\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0zetapi/2 is your new zea

anonymous
 3 years 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

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

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0a 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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