Q25·Section II (Mathematics)·25 of 145∫01x−1/3dx=32\int_0^1x^{-1/3}dx=\tfrac32∫01x−1/3dx=23The Lebesgue integral of the function f:[0,1]→Rf:[0,1]\to\mathbb{R}f:[0,1]→R defined by f(x)=1/x3f(x)=1/\sqrt[3]xf(x)=1/3x if 0<x≤10<x\le10<x≤1, f(0)=0f(0)=0f(0)=0 is: OptionsA34\tfrac3443B23\tfrac2332C32\tfrac3223D43\tfrac4334Pick an option to check your answer.Show answerShow solution Report a problem with this question← Q24All questionsQ26 →Keyboard: A–D answer · S solution · ←/→ previous/next