If ∫ 0 1 1 3 + x + 1 + x d x = a + b 2 + c 3 , where a , b , c are rational numbers, then 2 a + 3 b - 4 c is equal to :

Step-by-step solution

Let, I = ∫ 0 1 1 3 + x + 1 + x d x ⇒ I = ∫ 0 1 3 + x - 1 + x ( 3 + x ) - ( 1 + x ) d x ⇒ I = 1 2 ∫ 0 1 3 + x d x - ∫ 0 1 ( 1 + x ) d x ⇒ I = 1 2 2 ( 3 + x ) 3 2 3 - 2 ( 1 + x ) 3 2 3 0 1 ⇒ I = 1 2 2 ( 3 + 1 ) 3 2 3 - 2 ( 1 + 1 ) 3 2 3 - 2 ( 3 ) 3 2 3 - 2 ( 1 ) 3 2 3 ⇒ I = 1 2 2 × 8 3 - 4 2 3 - 2 × 3 3 3 + 2 3 ⇒ I = 1 2 6 - 4 2 3 - 2 3 ⇒ I = 3 - 3 - 2 3 2 = a + b 2 + c 3 ⇒ a = 3 , b = - 2 3 , c = - 1 ⇒ 2 a + 3 b - 4 c = 6 - 2 + 4 = 8