jee-main 2024 Q89

jee-main · India · session2_05apr_shift2 Areas by integration
Let $y = y ( x )$ be the solution of the differential equation $\frac { \mathrm { d } y } { \mathrm {~d} x } + \frac { 2 x } { \left( 1 + x ^ { 2 } \right) ^ { 2 } } y = x \mathrm { e } ^ { \frac { 1 } { \left( 1 + x ^ { 2 } \right) } } ; y ( 0 ) = 0$. Then the area enclosed by the curve $f ( x ) = y ( x ) \mathrm { e } ^ { - \frac { 1 } { \left( 1 + x ^ { 2 } \right) } }$ and the line $y - x = 4$ is $\_\_\_\_$
Let $y = y ( x )$ be the solution of the differential equation $\frac { \mathrm { d } y } { \mathrm {~d} x } + \frac { 2 x } { \left( 1 + x ^ { 2 } \right) ^ { 2 } } y = x \mathrm { e } ^ { \frac { 1 } { \left( 1 + x ^ { 2 } \right) } } ; y ( 0 ) = 0$. Then the area enclosed by the curve $f ( x ) = y ( x ) \mathrm { e } ^ { - \frac { 1 } { \left( 1 + x ^ { 2 } \right) } }$ and the line $y - x = 4$ is $\_\_\_\_$