The value of $\int_{1}^{2}\left\lbrack f\left\{ g(x) \right\} \right\rbrack^{- 1}\ f^{'}\left\{ g(x) \right\} g^{'}(x)dx$, where $g(1) = g(2)$ is equal to

Question

Dr. Abubacker · Accepted Answer

(c)
Let $I = \int_{1}^{2}{\lbrack f\left\{ \left( g(x) ight\} ightbrack^{- 1}f^{'}\left\{ g(x) ight\} g^{'}(x)dx}$
Put $f\left( g(x) ight) = t\  \Rightarrow f^{'}\left( g(x) ight)g^{'}(x)dx = dt$
$	herefore I = \int_{1}^{2}t^{- 1}dt = \lbrack\log{f\left\{ g(x) ight\}\left. \ \  ightbrack_{1}^{2}}$
$= \log{f\{ g(2)\} - log\ f\{ g(1)\} = 0}$
$\lbrack\because g(2) = g(1)brack$

Q. The value of $\int_{1}^{2}\left\lbrack f\left\{ g(x) \right\} \right\rbrack^{- 1}\ f^{'}\left\{ g(x) \right\} g^{'}(x)dx$, where $g(1) = g(2)$ is equal to

Explanation

Reviewed by Dr. Abubacker

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