strange calc problem

[vassago]

f(x) is continuous and decreasing on the closed interval 6<=x<=13, f(6)=7 and f(13)=4 and the integral f(x)dx = 42.943461 on [6,13] so find the integral f^-1(x)dx on [4,7]?

 

[royhaas]

Evaluate \(\displaystyle \int_4^7 f^{-1}(x)dx\) by making the substitution \(\displaystyle x=f(u)\).

 

[courteous]

Evaluate \(\displaystyle \int_4^7 f^{-1}(x)dx\) by making the substitution \(\displaystyle x=f(u)\).

(In hope of not hijacking the thread.)
\(\displaystyle \int_4^7 f^{-1}(x)dx = (\rightarrow x=f(u) \text{ and } dx=df(u) \Rightarrow) = \int_4^7 f^{-1}(f(u))df(u) = ?\) Stuck ... (not even sure about made substitution (and integral boundaries) ), what next? How would you use given information, that f(x)dx = 42.943461?

 

[Deleted member 4993]

f(x) is continuous and decreasing on the closed interval 6<=x<=13, f(6)=7 and f(13)=4 and the integral f(x)dx = 42.943461 on [6,13] so find the integral f^-1(x)dx on [4,7]?

\(\displaystyle f^{-1}(4) \, = \, 13\)

and

\(\displaystyle f^{-1}(7) \, = \, 6\)

then

\(\displaystyle \int_4^7f^{-1}(x)dx\)

\(\displaystyle = \, \int_{13}^6f^{-1}(f(u))f'(u)du\)

\(\displaystyle = \, \int_{13}^6u\cdot f'(u)du\)

and then continue....