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For the continuous random variable $X$ with pdf $f()$,

$$f(x)=\begin{cases}cx(x+1)&\text{for}&3

Find $c$.

Also,

For the previous question, find expectation of $X$.

I need help trying to understand this question so I can properly study for my test this Friday.

Thanks

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    What do you mean by "for 3"? Your question lacks info (the domain of the pdf) and cannot be answered properly. To find $c$ use the equality $\int f(x)dx=1$. To find the expectation use $\mathbb EX=\int xf(x)dx$. – 2017-02-07
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    I'm going to ignore the '3' and assume it's "for $x \in (0,1).$" Then this is a Beta distribution, which you can investigate in Wikipedia. But please do edit your Question so it makes sense. – 2017-02-07
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    @drhab It was a formatting issue: If you don't place them inside MathJax delimiters, the system interprets $<$ as the start of of HTML tag and 'eats' the rest of the line. – 2017-02-08
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    barbella321 , take a little time to peruse the [MathJax basic tutorial and quick reference](http://meta.math.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference) . It will help you properly format future posts. – 2017-02-08

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a) It is a property of a continuous random variable that the integral of the pdf over the support equals 1.   Find the value of $c$ that makes this so. $$1 = \int_3^5 cx(x+1) \operatorname d x$$

b) Expectation of a continuous random variable has a definition; so use it.

$$\mathsf E(X) = \int_3^5 x\cdot cx(x+1)\operatorname d x$$

Everything else is just integration of polynomials.