MHT CET202527 Apr 2025Evening ShiftMathematicsDefinite IntegrationActual
₀^ 6 (2+3 x^2 ) 3 x d x=
Options
- A2 9 + ^2 36
- B4 9 + ^2 36
- C2 9 - ^2 36
- D4 9 - ^2 36
Correct answer
A. 2 9 + ^2 36
Step-by-step solution
Evaluate the definite integral: ₀^ 6 (2+3x^2) 3x ,dx Apply integration by parts with u = 2+3x^2 and dv = 3x ,dx , yielding du = 6x ,dx and v = 1 3 3x : I = 1 3 (2+3x^2) 3x - 2 x 3x ,dx Apply integration by parts again to the remaining integral with u = x and dv = 3x ,dx , giving du = dx and v = - 1 3 3x : x 3x ,dx = - 1 3 x 3x + 1 9 3x Substitute back into the expression for I : I = 1 3 (2+3x^2) 3x + 2 3 x 3x - 2 9 3x Evaluate the definite integral from 0 to 6 : At x = 6 : 2 = 1 , 2 = 0 1 3 (2+3( 6 )^2)(1) + 2 3 (