What is the value of C so that the following function is a probability density function? f(x)=Cx^9.5 --> see description

Question

f(x)=Cx^9.5 if 0≤x≤2, 0 otherwise

Answer 1

So in order for it to be a probability density function, the area has to be equal to 1, so let's integrate this function for 0 to 2 and set it equal to 1:

f(x)=Cx^9.5, F(x)=Cx^10.5/10.5,

Now we plug in 2 for x and subtract when we plug in 1 for x

F(2)-F(1)=1; C/10.5*2^10.5-C/10.5*1^10.5=1, C(2^10.5-1)=10.5

C=10.5/(2^10.5-1) or C=10.5/1447.15

Answer 2

Well the integral from 0 to 2 of f(x) has to be 1.

If you integrate Cx^9.5 you get 10.5C x^10.5

Between limits 2 and 0 this evaluates to 10.5 C 2^10.5

This is 1, giving C = 1 / (10.5 x 2^10.5)