2

If I have this function :$$\ f(x)=x^3+x $$ how can I calculate integral from inverse $$\int_0^1 f^{-1}(x)\,dx $$ ? I understand that in order do solve this problem I have to take $$\ x=f(t) $$ $$\ dx=f'(t) dt $$ and in the end I'll have $$\int t*f'(t)dt $$ but I don't understand how I find the lower and upper bound of this integral without calculating the actual inverse of the function.

A.Γ.
  • 29,518
Lola
  • 1,601
  • 1
  • 8
  • 19
  • 1
    Integration limits are from $a=f^{-1}(0)$ to $b=f^{-1}(1)$, that is, the solutions to $f(a)=0$ and $f(b)=1$. The first one is easily $0$, the second one is nasty. – A.Γ. Apr 29 '17 at 12:39
  • Thank you,I understand how to find the bounds now,the second solution is kind of weird but I doubt they wanted to write the integral like this in the book.I think they wanted to write from 0 to 2 since the last integral is from 0 to 1 – Lola Apr 29 '17 at 14:20

2 Answers2

4

There is a graphical proof of Young's inequality... the picture is:

enter image description here

And in case $b=f(a)$ you get a formula for one integral by subtracting the other from the area of a rectangle.

(When $b \ne f(a)$, the red area sticks out of the rectangle, which is where you get the "inequality" in the proof of Young's inequality.)

GEdgar
  • 111,679
1

You put $t=f (x)$ or $x=g (t) $ where $g=f^{-1} $.

then $$\int_0^1f (x)dx=\int_{f (0)}^{f (1)} tg'(t)dt $$

with

$$g'(t)=\frac {1}{f'(g (t))} $$

hamam_Abdallah
  • 62,951
  • I got at this point but I don't know how to find the upper and lower bound of the integral,knowing that my first integral is from 0 to 1(forgot to put it first I edited it now) – Lola Apr 29 '17 at 11:03
  • @Lola is it okay now – hamam_Abdallah Apr 29 '17 at 13:38
  • Well in the book it says that the second integral is from 0 to 1 and I don't get those bounds if I calculate f(0) and f(1) – Lola Apr 29 '17 at 14:13