Multiplex! for JEE

$L=\displaystyle \lim _{ x\to 0 } \left( \int _{ 0 }^{ 1 } (by+a(1-y))^{ x }.dy \right) ^{ \frac { 1 }{ x } }\\ I=\displaystyle \int _{ 0 }^{ 1 } (by+a(1-y))^{ x }.dy\\ Let\quad by+a(1-y)=t\quad ,\quad (b-a)dy=dt\\ (need\quad not\quad compute\quad the\quad bounds)\\ I=\displaystyle \int { \frac { { t }^{ x } }{ (b-a) } } dt\quad =\quad \frac { { t }^{ x+1 } }{ (x+1)(b-a) } \\ I=\displaystyle \left[ \frac { ({ by+a(1-y)) }^{ x+1 } }{ (x+1)(b-a) } \right] \quad computed\quad at\quad y=1\quad and\quad y=0\\ I=\displaystyle \frac { { b }^{ x+1 }-{ a }^{ x+1 } }{ (x+1)(b-a) } \\ L=\displaystyle \lim _{ x\to 0 } \left( \frac { { b }^{ x+1 }-{ a }^{ x+1 } }{ (x+1)(b-a) } \right) ^{ \frac { 1 }{ x } }\\ \displaystyle In(L)=\lim _{ x\to 0 } \frac { In\left( \frac { { b }^{ x+1 }-{ a }^{ x+1 } }{ (x+1)(b-a) } \right) }{ x } \\ As\quad x\quad approaches\quad 0, denominator\quad and\quad numerator\quad approach\quad 0.\\ Hence,\quad we\quad can\quad use\quad L'hopital's\quad rule.\\ In(L)=\displaystyle \lim _{ x\to 0 } \left( \frac { { b }^{ x+1 }In(b)-{ a }^{ x+1 }In(a) }{ (x+1)(b-a) } -\frac { 1 }{ (x+1) } \right) \\ In(L)=\displaystyle \frac { 1 }{ (b-a) } In\frac { { b }^{ b } }{ { a }^{ a } } -1\\ L=\boxed { \boxed {\displaystyle \frac { 1 }{ e } { \left( \frac { { b }^{ b } }{ { a }^{ a } } \right) }^{ \frac { 1 }{ (b-a) } } } }$