Let f(x) = a [x+1] + b [2x-1] , where a, b R and [t] denotes the greatest integer less than or equal to t . If _ x 1 f(x) exists and ₀^2 f(x) dx = 6 , then the value of _ -1 ^1 f(x) dx is equal to

Step-by-step solution

Given f(x) = a [x+1] + b [2x-1] . For the limit _ x 1 f(x) to exist, the left-hand limit (LHL) and right-hand limit (RHL) at x = 1 must be equal. LHL at x = 1 ( x 1^- ): [x+1] 1 and [2x-1] 0 LHL = a(1) + b(0) = a RHL at x = 1 ( x 1^+ ): [x+1] 2 and [2x-1] 1 RHL = a(2) + b(1) = 2a + b Equating LHL and RHL: a = 2a + b b = -a Now, evaluate ₀^2 f(x) dx = a ₀^2 [x+1] dx - a ₀^2 [2x-1] dx . ₀^2 [x+1] dx = ₀^1 1 dx + ₁^2 2 dx = 1 + 2 = 3 ₀^2 [2x-1] dx = ₀^ 0.5 (-1) dx + _ 0.5 ^1 0 dx + ₁^ 1.5 1 dx + _ 1.5 ^2 2 dx = -0.5 +