Question 364992
The integral of the density function from {{{x=0.9}}} to {{{x=1.1}}} must equal {{{1}}}.
{{{int(k(x-0.9)(1.1-x),dx,0.9,1.1)=1}}}

{{{k*int((-x^2+2x-(99/100)),dx,0.9,1.1)=1}}}
{{{-((1.1)^3-(0.9)^3)/3+(1.1)^2-(0.9)^2-(99/100)(1.1-0.9)=1/k}}}
{{{-((1.1)^3-(0.9)^3)/3+(1.1)^2-(0.9)^2-(99/100)(1.1-0.9)=1/k}}}
{{{-602/3000+1200/3000-594/3000=1/k}}}
{{{1/k=4/3000}}}
{{{highlight(k=750)}}}
.
.
{{{graph(300,300,0.9,1.1,-1,8,750(x-0.9)(1.1-x))}}}
.
.
The left endpoint is {{{x=0.9}}}, the right endpoint is {{{x=1.1}}}.
The peak of {{{7.5}}} occurs when {{{x=1}}}.
.
.
Now that you have {{{k}}} you can solve for the mean and the variance.
{{{mu=int(x*p(x),dx)=k*int((-x^3+2x^2-(99/100)x),dx,0.9,1.1)}}}

{{{mu/k=-0.25((1.1)^4-0.9^4)+(2/3)((1.1)^3-0.9^3)-(99/2000)(1.1-0.9)}}}
{{{highlight(mu=1)}}}
.
.
.
{{{V=int((x-1)^2*p(x),dx)=k*int((-x^4+4x^3-(599/100)x^2+(398/100)x-99/100),dx,0.9,1.1)}}}
{{{V/k=-(1/5)((1.1)^5-0.9^5)+(599/300)((1.1)^3-0.9^3)+(199/100)(1.1^2-0.9^2)-0.99(1.1-0.9)}}}
{{{V/k=8/3000000}}}
{{{highlight(V=1/500)}}}