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JonaskBest ResponseYou've already chosen the best response.0
prove that there exist a \[\xi\] in \[0 \le x \le \pi\] such that \[\huge \int\limits _0^\pi e^{x}\cos x dx =\sin \xi \]
 one year ago

amistre64Best ResponseYou've already chosen the best response.0
integrate the left side to determine its value
 one year ago

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

amistre64Best ResponseYou've already chosen the best response.0
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
 one year ago

satellite73Best ResponseYou've already chosen the best response.0
my guess is that you do not have to compute anything (but i could be wrong)
 one year ago

amistre64Best ResponseYou've already chosen the best response.0
..... but .... i like computing :)
 one year ago

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

amistre64Best ResponseYou've already chosen the best response.0
cosx changes sign in the interval, e^x stays positive tho so we can let g(x)=e^x
 one year ago

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

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

amistre64Best 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
 one year ago

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

satellite73Best ResponseYou've already chosen the best response.0
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
 one year ago

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

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

satellite73Best ResponseYou've already chosen the best response.0
\(C=\int_0^{\pi}=e^{x}dx\)
 one year ago

satellite73Best ResponseYou've already chosen the best response.0
i mean \[C=\int_0^{\pi}e^{x}dx\]
 one year ago

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

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

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

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

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

satellite73Best ResponseYou've already chosen the best response.0
\[\int_{\frac{\pi}{2}}^{\frac{\pi}{2}}e^{x\frac{\pi}{2}}\sin(x)dx\]
 one year ago

satellite73Best ResponseYou've already chosen the best response.0
but unfortunately now the interval changed...
 one year ago

satellite73Best ResponseYou've already chosen the best response.0
yeah but how do you know \[0<\xi<\pi\]??
 one year ago

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

satellite73Best ResponseYou've already chosen the best response.0
then you don't have sine anymore
 one year ago

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

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

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

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

satellite73Best ResponseYou've already chosen the best response.0
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
 one year ago

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

electrokidBest 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
 one year ago

electrokidBest ResponseYou've already chosen the best response.1
a great man once said... math becomes very meaningful when you keep the mathematical meanings out!!! it will lead to logic :) (me)
 one year ago
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