JEE Main202228 Jul 2022Evening ShiftMathematicsDefinite IntegrationActual
Let I n x = ∫ 0 x 1 t 2 + 5 n d t , n = 1 , 2 , 3 , … . Then
Options
- A50 I 6 - 9 I 5 = x I 5 '
- B50 I 6 - 11 I 5 = x I 5 '
- C50 I 6 - 9 I 5 = I 5 '
- D50 I 6 - 11 I 5 = I 5 '
Correct answer
A. 50 I 6 - 9 I 5 = x I 5 '
Step-by-step solution
Given, I n x = ∫ 0 x d t t 2 + 5 n Applying integration by parts we get, ⇒ I n x = t t 2 + 5 n 0 x - ∫ 0 x n t 2 + 5 - n - 1 · 2 t 2 ⇒ I n x = x x 2 + 5 n + 2 n ∫ 0 x t 2 t 2 + 5 n + 1 d t ⇒ I n x = x x 2 + 5 n + 2 n ∫ 0 x t 2 + 5 - 5 t 2 + 5 n + 1 d t ⇒ I n x = x x 2 + 5 n + 2 n ∫ 0 x d t t 2 + 5 n - 10 n ∫ 0 x d t t 2 + 5 n + 1 ⇒ I n x = x x 2 + 5 n + 2 n I n x - 10 n I n + 1 x ⇒ 10 n I n + 1 x + 1 - 2 n I n x = x x 2 + 5 n ⇒ 10 n