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$Using\quad Integration\quad to\quad get\quad { A }_{ 1 }\quad ,\quad { A }_{ 2 }\\ { A }_{ 1 }=\int _{ 0 }^{ 1 }{ 2x{ e }^{ x }dx } \quad \longrightarrow \quad (1)\\ { A }_{ 2 }=\int _{ 0 }^{ 1 }{ { e }^{ \sin { (x) } }\sin { (2x) } } dx\\ \because \quad \sin { (2x) } =2\sin { (x)\cos { (x) } } \\ \therefore { \quad A }_{ 2 }=\int _{ 0 }^{ 1 }{ 2 } { e }^{ \sin { (x) } }\sin { (x)\cos { (x) } } dx\\ Let\quad u=sin(x)\\ Change\quad the\quad boundaries\quad in\quad terms\quad of\quad u\\ x=0\quad \Longrightarrow \quad u=\sin { (0)=0 } \\ x=1\quad \Longrightarrow \quad u=\sin { (1) } \\ du=\cos { (x) } dx\\ \\ \therefore \quad { A }_{ 2 }=\int _{ 0 }^{ \sin { (1) } }{ 2u{ e }^{ u } } du\quad \longrightarrow \quad (2)\\ From\quad (1)\quad ,\quad (2)\\ \int _{ 0 }^{ 1 }{ 2x{ e }^{ x }dx } >\int _{ 0 }^{ \sin { (1) } }{ 2u{ e }^{ u } } du\\ \\ \boxed { \therefore { A }_{ 1 }{ \quad >\quad A }_{ 2 } }$