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?


ANSWER

$$0$$


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
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