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integrate the left side to determine its value

i used by parts
and i have
\[2I=e^{-x}(\sin x-\cos x)|_0^\pi\]

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

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

\[0\le \xi-\pi/2\le \pi\]

\[0\le \xi+\pi/2\le \pi\]

zeta-pi/2 is your new zea