Let XXX have PDF f(x)=kx3f(x) = kx^3f(x)=kx3 for 0≤x≤20 \le x \le 20≤x≤2. Find kkk such that ∫02f(x)dx=1\int_0^2 f(x)dx = 1∫02f(x)dx=1.
k=14k = \frac{1}{4}k=41
k=18k = \frac{1}{8}k=81
k=116k = \frac{1}{16}k=161
k=12k = \frac{1}{2}k=21