MVT for integrals

- anonymous

MVT for integrals

- katieb

See more answers at brainly.com

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions

- 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.