MVT for integrals

- anonymous

MVT for integrals

- Stacey Warren - Expert brainly.com

Hey! We 've verified this expert answer for you, click below to unlock the details :)

- jamiebookeater

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

- anonymous

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

- amistre64

integrate the left side to determine its value

- anonymous

##### 1 Attachment

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- anonymous

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

- amistre64

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

- anonymous

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

- amistre64

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

- anonymous

okay did you have a look at the MVTS attached

- amistre64

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

- anonymous

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

- anonymous

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

- amistre64

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

- anonymous

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

- anonymous

i mean def 2

- anonymous

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

- anonymous

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

- anonymous

exacly

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

obviusly we know the primitive of cos is sin

- anonymous

yes

- anonymous

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

- anonymous

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

- anonymous

oh good idea!

- anonymous

are you talking about the equation or the last conlution

- anonymous

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

- anonymous

really good idea

- anonymous

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

- anonymous

yep

- anonymous

but unfortunately now the interval changed...

- anonymous

so what? I = I

- anonymous

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

- anonymous

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

- anonymous

then you don't have sine anymore

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

zeta-pi/2 is your new zea

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

Looking for something else?

Not the answer you are looking for? Search for more explanations.