Can someone just get me on the right track for this one, probability density related standard deviation and V?

Consider the function p1(x), −b x b, which is given by


p1(x) =


(c/b)x+c if − b x 0





(-c/b)x + c if 0 x b


Find the variance and the standard deviation for p1(x).





do i first find the mean? and then find the rest?

Can someone just get me on the right track for this one, probability density related standard deviation and V?
You do have to find the mean first, as you need it to calculate variance.





E(X) = Integral(-b to b)xp1(x)dx





E(X) = Integral(-b to 0)x(cx/b+c)dx + Integral(0 to b)x(-cx/b+c)dx





An antiderivative of x(cx/b +c) is cx^3/(3b) + cx²/2


An antiderivative of x(-cx/b +c) is -cx^3/(3b) + cx²/2


E(X) = [0 - (-cb^3/(3b) + cb²/2)] + [-cb^3/(3b) + cb²/2 -0] = 0


The mean is 0.





Var(X) = Integral(-b to b)(x-E(x))²p1(x)dx = Integral(-b to b)x²p1(x)dx


Var(X) = Integral(-b to 0)x²(cx/b+c)dx + Integral(0 to b)x²(-cx/b+c)(x)dx





An antiderivative of x²(cx/b +c) is cx^4/(4b) + cx^3/3


An antiderivative of x²(-cx/b +c) is -cx^4/(4b) + cx^3/3


Var(X) = [0 - (cb^4/(4b) - cb^3/3)] + [-cb^4/(4b) + cb^3/3 -0]





Var(X) = -cb^3/2 + 2cb^3/3 = cb^3/6





the standard deviation is σ(x) = √(Var(X)) = √(cb^3/6)