BITSAT Mathematics Question (2021) — Solution

Step-by-step solution

We have, _ 0 ^ 1 5+4 x d x = _ 0 ^ 1 5+4 ( 1- ^ 2 x 2 1+ ^ 2 x 2 ) d x = _ 0 ^ 1+ ^ 2 x 2 5 (1+ ^ 2 x 2 )+4 (1- ^ 2 x 2 ) d x = _ 0 ^ 1+ ^ 2 x 2 9+ ^ 2 x 2 dx = _ 0 ^ ^ 2 x 2 9+ ^ 2 x 2 dx Let x 2 =t 1 2 ^ 2 x 2 d x=d t Also, x=0 t=0 and x= t= I = _ 0 ^ dt 9+t^ 2 I =2 _ 0 ^ dt 9+t^ 2 I = 2 3 [ ^ -1 t 3 ]_ 0 ^ = 2 3 [ ^ -1 - ^ -1 0 ] = 2 3 ( 2 -0 )= 3