Let n be a positive integer. For a real number x , let [x] denote the largest integer not exceeding x and x =x-[x] . Then ₁^ n+1 ( x )^ [x] [x] d x is equal to

Step-by-step solution

₁^ n+1 x ^ [x] [x] d x= ₁² x ^ [x] [x] d x+ ₂³ x ^ [x] [x] d x+ .+ ₁^ n+1 x ^ [x] [x] d x= _ r=1 ^ n _ r ^ r+1 x ^ [x] [x] d x= _ r=1 ^ n _ r ^ r+1 (x-r)^ r r d x= _ r=1 ^ n [ (x-r)^ r+1 r(r+1) ]_ r ^ r+1 = _ r=1 ^ n 1 r(r+1) = n n+1