Mathematics

# The value of $\displaystyle \int_{1}^{2}\left [ f\left \{ g\left ( x \right ) \right \} \right ]^{-1}.{f}'\left \{ g\left ( x \right ) \right \}.{g}'\left ( x \right )dx,$ where $\displaystyle g\left ( 1 \right )=g\left ( 2 \right ),$ is equal to?

$0$

##### SOLUTION
Let $I=\int _{ 1 }^{ 2 } \left[ f\left\{ g\left( x \right) \right\} \right] ^{ -1 }.{ f }'\left\{ g\left( x \right) \right\} .{ g }'\left( x \right) dx$

Substitute $\left[ f\left\{ g\left( x \right) \right\} \right] =t\Rightarrow { f }'\left\{ g\left( x \right) \right\} .{ g }'\left( x \right) dx=dt$

$\displaystyle \therefore I=\int _{ f\left( g\left( 1 \right) \right) }^{ f\left( g\left( 2 \right) \right) }{ tdt } ={ \left[ \frac { { t }^{ 2 } }{ 2 } \right] }_{ f\left( g\left( 1 \right) \right) }^{ f\left( g\left( 2 \right) \right) }=0\\ \because g\left( 1 \right) =g\left( 2 \right)$

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Single Correct Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 86

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