given-the-pdf-fx-3x2-e-x3-for-x-0-and-fx-0-for-x

Question

Given the PDF f(x) = $$3x^2$$ $$e^{-x^3}$$ for x ≥ 0 and f(x) = 0 for x

Accepted Answer

F(x) = 1 - $$e^{-x^3}$$ for x ≥ 0 (and 0 for x \u003c 0)

Answer

F(x) = $$-e^{-x^3}$$ for x ≥ 0 and undefined for x \u003c 0, because the constant of integration is neglected

Answer

F(x) = 1 - $$e^{x^3}$$ for x ≥ 0, since integrating the negative exponent flips the sign, and 0 for x \u003c 0

Answer

F(x) = $$e^{-x^3}$$ for x ≥ 0 and 1 for x \u003c 0 because the integral of the density is the exponential evaluated at the negative cube

Given the PDF f(x) = 3x^2 e^{-x^3} for x ≥ 0 and f(x) = 0 for x < 0, compute the cumulative distribution function F(x) for x ≥ 0.