
Kenneth S. answered 11/16/16
Tutor
4.8
(62)
Expert Help in Algebra/Trig/(Pre)calculus to Guarantee Success in 2018
For a function to be a PDF, its definite integral over is domain must = 1.
The indefinite integral for the interval [-1,0) is (-k/2)x2 and the associated definite integral value is k/2.
The definite integral for the interval [0,1] is k.
So we have k +k/2 = 1 which gives k = 2/3.
you can do iii.
Sasha B.
11/19/16