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We assume that you want to find $E(X)$, and the variance of $X$. The expectation of $X$ is $\int_{-\infty}^\infty xf_X(x)\,dx$. In our case, this is $\int_0^1 (x)(x)\,dx+\int_1^2 (x)(2-x)\,dx$. We can calculate, but by symmetry of the density function (draw a picture) the answer is $1$. For the variance, we need to calculate $E(X^2)-(E(X))^2$. We have ...


I assume the following distribution: $f(x)=\begin{cases} x, \ \ \text{if} \ \ 0 < x <1 \\ 2-x, \ \ \text{if} \ \ 1 \leq x <2 \\ 0, \ \ \text{elsewhere} \end{cases}$ $E(x)=\int_0^1 (x \cdot x) \ dx+\int_1^2 x \cdot (2-x) \ dx$. Now calculate E(x). greetings, calculus

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