Given the dataset which is formed by the independent random variables $X_1, X_2, ..., X_N$. Suppose that every $X_i$ has a cumulative distribution function $F$ where:
$$ F(x) = \begin{cases} 0 & \text{if $x<0$} \\ \left(\frac{x}{\theta}\right)^2 & \text{if $0\leq x \leq \theta$}\\ 1 & \text{if $x>\theta$ } \end{cases}$$
where $\theta>0$ is unknown. Consider the following estimator for $\theta$ $$T = a(X_1+X_2+ ... +X_n) + b$$ For which $a$ and $b$ is $T$ an unbiased estimator of $\theta$?
I know that for an unbiased estimator $T$ it must hold that $\mathbb{E}(T) = \theta$. I tried calculating $\mathbb{E}(T)$ by taking the derivative of $F(x)$ (define $F'(x) = f(x))$. And I thought that: $$\mathbb{E}(T) = \int_0^\theta x\ f(x)\ dx$$ But I can't seem to reach a conclusion on what $a$ and $b$ should be. How do I proceed? Or did I make a mistake?
Update
Is it correct to say the following: $$\int_0^\theta x\ f(x)\ dx = \int_0^\theta \frac{2x^2}{\theta^2}dx = \frac{2}{3}\theta$$ So $$\mathbb{E}(T) = \mathbb{E}(a(X_1+X_2+...+X_n)+b) = (a)(n)\mathbb{E}(X_1)+b$$ So therefore $b=0$ and $a=\frac{3}{2n}$?
The $X_i$ are iid and with linearity of expectation and symmetry you can find: $$\mathbb ET=an\mathbb EX_1+b$$
Find $\mathbb EX_1$ (so not $\mathbb ET$ directly as you write in your question) by the way you suggested:$$\mathbb{E}X_1 = \int_0^\theta x\ f(x)\ dx$$ where $f$ denotes the derivative of $F$ hence functions as PDF.
You will find an expression in $\theta$. Let's say $\mathbb EX_1=g(\theta)$.
Then to be found are values of $a,b$ such that $$\theta=ang(\theta)+b$$
So the answer is a work out of the set: $$\{\langle a,b\rangle\mid \theta=ang(\theta)+b\}$$