Let J = ₀^ /4 2x ^2 x x (3 ^4 x - 6 ^2 x + 2) d x . Then the value of 15J is equal to :

Step-by-step solution

Let J = ₀^ /4 x 2 ^2 x x (3 ^4 x - 6 ^2 x + 2) d x . Apply integration by parts taking u = x and d v = 2 ^2 x x (3 ^4 x - 6 ^2 x + 2) d x . To find v , let y = ^2 x , then d y = 2 ^2 x x d x . v = (3y^2 - 6y + 2) d y = y^3 - 3y^2 + 2y = ^6 x - 3 ^4 x + 2 ^2 x . Now, evaluate the boundary terms: v(0) = 1 - 3 + 2 = 0 v( /4) = ( 2 )^6 - 3( 2 )^4 + 2( 2 )^2 = 8 - 12 + 4 = 0 Thus, the boundary term [x v(x)]₀^ /4 vanishes. The remaining integral is: J = - ₀^ /4 v(x) d x = - ₀^ /4 ( ^6 x - 3 ^4 x + 2 ^2 x) d x Factor out