Question

MVT for integrals

Answer 1

prove that there exist a \[\xi\]
in \[0 \le x \le \pi\] such that
\[\huge \int\limits _0^\pi e^{-x}\cos x dx =\sin \xi \]

Answer 2

integrate the left side to determine its value

Answer 3

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

Answer 4

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

Answer 5

my guess is that you do not have to compute anything (but i could be wrong)

Answer 6

..... but .... i like computing :)

Answer 7

okay did you have a look at the MVTS attached

Answer 8

cosx changes sign in the interval, e^-x stays positive tho so we can let g(x)=e^-x

Answer 9

plugging in the boundaries gives
\[\frac{1}{2}(e^{-\pi}+1)\le 2(1/2)\le 1\]

Answer 10

i agree with @satellite73 in mean value teorems theres very little comp.

Answer 11

\[f(E)=\frac{1}{b-a}\int_{a}^{b}f(x)dx\] might be helpful? or i mightbe barking up the wrong tree

Answer 12

well since we have two functions i figured that we might need a defination with f and g

Answer 13

i mean def 2

Answer 14

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

Answer 15

ane other helpful toll
\[m(b-a) \le \int\limits _a^b f(x)dx \le M(b-a)\]

Answer 16

exacly

Answer 17

i assume C
is\[e^{- \xi}\]

Answer 18

\(C=\int_0^{\pi}=e^{-x}dx\)

Answer 19

i mean
\[C=\int_0^{\pi}e^{-x}dx\]

Answer 20

obviusly we know the primitive of cos is sin

Answer 21

yes

Answer 22

how about replacing x with x=u-pi/2?

Answer 23

then cos becomes sin, exp(-x) can still be taken as g(x)...

Answer 24

oh good idea!

Answer 25

are you talking about the equation or the last conlution

Answer 26

in the original eq.. then follow the same steps as above...

Answer 27

really good idea

Answer 28

\[\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}e^{x-\frac{\pi}{2}}\sin(x)dx\]

Answer 29

yep

Answer 30

but unfortunately now the interval changed...

Answer 31

so what? I = I

Answer 32

yeah but how do you know
\[0<\xi<\pi\]??

Answer 33

then, zeta lies in [-pi/2,pi/2]
then translate the co-ordinates back to "x"

Answer 34

then you don't have sine anymore

Answer 35

you are looking at the same space but you have repositioned yourself at a different point is space...

Answer 36

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

Answer 37

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

Answer 38

zeta-pi/2 is your new zea

Answer 39

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

Answer 40

fine name the one obtained in "UV" plane as "eta"
then that same "eta" is "zeta" in "XY" plane...
function did not change

Answer 41

\[
-{\pi\over2}\le\eta\le{\pi\over2},\qquad-{\pi\over2}\le u\le{\pi\over2}\\
0\le\eta+{\pi\over2}\le\pi,\qquad0\le u+{\pi\over2}\le\pi\\
0\le\zeta\pi,\qquad0\le x\le\pi
\]
we had

Answer 42

a great man once said...
math becomes very meaningful when you keep the mathematical meanings out!!! it will lead to logic
:)
(me)