$\int\limits_1^4(\{x\})^{[x]}dx$, where [.] denotes the greater integer function and {.} denotes fraction part, is equal to: 

$\frac{13}{12}$

$\int\limits_1^4(\{x\})^{[x]}dx=\int\limits_1^2(x-1)dx+\int\limits_2^3(x-2)^2dx+\int\limits_3^4(x-3)^3dx$

$=\int\limits_0^1tdt+\int\limits_0^1t^2dt+\int\limits_0^1t^3dx=[\frac{t^2}{2}+\frac{t^3}{3}+\frac{t^4}{4}]_0^1=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}=\frac{13}{12}$