Let f(x) = a ^4 x - b ^3 x + (a+1) ^2 x - b x + 1 , where a and b are real constants. If ₀^ /4 f(x) d x = 0 and ₀^ /4 x f(x) d x = 3- 2 , then the value of a + b is equal to :

Step-by-step solution

First, factor the given function f(x) : f(x) = a ^2 x( ^2 x + 1) - b x( ^2 x + 1) + ( ^2 x + 1) f(x) = (a ^2 x - b x + 1) ^2 x Let F(x) = f(x) d x . Using the substitution t = x , we get: F(x) = a 3 ^3 x - b 2 ^2 x + x Given ₀^ /4 f(x) d x = 0 , we have F( /4) - F(0) = 0 . a 3 - b 2 + 1 = 0 2a - 3b + 6 = 0 For the second integral, apply integration by parts: ₀^ /4 x f(x) d x = [x F(x)]₀^ /4 - ₀^ /4 F(x) d x Since F( /4) = 0 and F(0) = 0 , the boundary term vanishes. Thus: - ₀^ /4 ( a 3 ^3 x - b 2 ^2 x + x ) d x = 3