Question

Evaluate \(\rm \int\limits_3^5 f^{-1}(x) dx \) if \(\rm f(x) = x^5+x+3\)

Answer 1

ganeshie... why yew do this?

Answer 2

https://soundcloud.com/dominick_d97/somebody-that-i-used-to-know-gotye-vs-a-sky-full-of-stars-coldplay

Answer 3

Umm..^ What if its for school? It's a learning site, isn't it? Or helping site?

Answer 4

haven't done Calc in a while lol

Answer 5

i like that song

Answer 6

Hes an adult XD

Answer 7

BOTH

Answer 8

the answer is 5% +6% %&#%$^%^%&#$&^*&*%^7

Answer 9

??? @thomaster @Preetha @jordanloveangel

Answer 10

Thank you @ikram002p

Answer 11

hmm interesting ...
i cant separate x and y , so whats in ur mind ?

Answer 12

Somebody claimed this can be done neatly using integration by parts.

I have worked using some other convoluted method...

Answer 13

so u have an answer hmm

Answer 14

yes, by using "geometry"

Answer 15

see
http://openstudy.com/study#/updates/52de9768e4b003c643a11d4f

Answer 16

Llalala!!!! @ganeshie8 may we plzzzzzz switch brains?

Answer 17

oh , thats ovc but algebraically hmmm

Answer 18

for sure yes lol
im still looking for a solution using IBP or any other method that doesn't require graphing

Answer 19

Yaya!!! Lol, Lets do it today and ummm 2:00pm at the surgery room, haha!!! I'm just kidding. Kk. I'll stop now.

Answer 20

u mean IVP ?

Answer 21

i think it would work but that would give a series something ... xD

Answer 22

I have been trying something like below :

say \(\large \rm y = f(x) = x^5+x+3\), I need to evaluate

\[\large \rm \int_3^5f^{-1}(x)~dx\]

substituting \(\rm x = y \) gives me

\[\large \rm \int_3^5f^{-1}(f(x))~dy\]

Answer 23

which is same as

\[\large \rm \int_3^5\!x~dy\]

Answer 24

haha maybe lets see if it gonna work
http://prntscr.com/560gqh

Answer 25

brb

Answer 26

geometrically thats same as :
\[5\times 1 - \int_0^1 y~ dx\]

\[5\times 1 - \int_0^1 x^5+x+3~~ dx\]


this looks a lot like IBP, but i was forced to use geometry for adding/subtracting areas

Answer 27

\[\begin{matrix}
u=f^{-1}(x)&&&dv=dx\\\\
du=\frac{1}{f'\left(f^{-1}(x)\right)}~dx&&&v=x
\end{matrix}\]

Answer 28

do i get \[\rm \int f^{-1}(x) dx = xf^{-1}(x) - \int x \cdot \dfrac{1}{f'\left(f^{-1}(x)\right)} dx\]

?

Answer 29

Yep, but there's still the matter of not "knowing" what \(f^{-1}(x)\) is. But if we did, we'd be able to figure out the integral from the start.

Answer 30

does the definite integral help ?

Answer 31

do i get \[\rm \int_3^5 f^{-1}(x) dx = xf^{-1}(x)\Bigg|_3^5 - \int_3^5 x \cdot \dfrac{1}{f'\left(f^{-1}(x)\right)} dx\]

?

Answer 32

I can certainly work \(\rm f^{-1}(5)\) and \(\rm f^{-1}(3)\) using f(x), not a problem. Evaluating the rightside integral looks like a pain though

Answer 33

Yes,
\[\int_a^b u~dv=uv\bigg|_a^b-\int_a^b v~du\]

Answer 34

what is IBP ?

Answer 35

integration by parts

Answer 36

-.-

Answer 37

For the second integral, you could consider a substitute \(u=f^{-1}(x)\), so that \(f(u)=x\) and \(f'(u)~du=dx\), if I have the right differentials.

Then
\[\int_3^5\frac{x}{f'(f^{-1}(x))}~dx=\int_{f(3)}^{f(5)}\frac{f(u)}{f'(f(u))}~f'(u)~du\]
if that helps...

Answer 38

lol.:P IBP.:D

Answer 39

its sounds like the same issue of morning problem !
nice

Answer 40

you meant this rihgt ? XD

\[\int_3^5\frac{x}{f'(f^{-1}(x))}~dx=\int_{f^{-1}(3)}^{f^{-1}(5)}\frac{f(u)}{f'(u)}~f'(u)~du \]

Answer 41

i think it works good , main idea is to get dx with respect to dy

Answer 42

Yes that's it, thanks

Answer 43

Beautiful !!! its IBP all the way xD thanks everyone!

Answer 44

I think we can generalize

Answer 45

\[\rm \int_a^b f^{-1}(x) dx = xf^{-1}(x)\Bigg|_a^b - \int_{f^{-1}(a)}^{f^{-1}(b)} \! f(x) dx\]

Answer 46

i hope it works..

Answer 47

but are u sure that 3,5 are valid in this integral xD

Answer 48

wym by valid ?

Answer 49

in the graph u showed me it sound like no shaded area with range of 3,5 :3

Answer 50

no no no forget it xD

Answer 51

here is the graph im using if you want
http://prntscr.com/560rnd

Answer 52

the shaded area is the area we're subtracing - the \(\int vdu\) component in IBP

Answer 53

:)

Answer 54

The general form works for at least a few examples. I tried it with \(f(x)=x^2\) over \([0,2]\). No counter-ex's yet.

Answer 55

I think this works
\[\int\limits_{3}^{5}f^{-1}(x) dx \\ u=f^{-1}(x) \\ f(u)=x \\ f'(u) du=dx \\ \int\limits_{0}^{1} u f'(u) du \\ \int\limits_{0}^{1} u (5u^4+1) du=\int\limits_{0}^{1}(5u^5+u) du\]

Answer 56

I didn't do integration by parts
but it seems like a sub works

Answer 57

@myininaya limits of integration?

Answer 58

Wow! that looks elegant xD

\[\begin{align} \int_a^b f^{-1}(x) dx &= xf^{-1}(x)\Bigg|_a^b - \int_{f^{-1}(a)}^{f^{-1}(b)} \! f(x) dx \\~\\
&=\int_{f^{-1}(a)}^{f^{-1}(b)} \!x f'(x) dx \\~\\
\end{align} \]

Answer 59

\[f^{-1}(5)=a \\ f(a)=5 \\ a^3+a+3=5 \\ a=1 \\ f^{-1}(5)=1 \\ f^{-1}(3)=b \\ f(b)=3 \\ b^3+b+3=3 \\ b=0 \\ f^{-1}(3)=0\]
that is how I got the limits @Princer_Jones

Answer 60

great..@myininaya ,.:)