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_problems/unit-03/B-expectation-of-a-continuous-random-variable/1.md
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---
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index: 1
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statement:
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-$$X(x) = \begin{cases} 1, &x \le 2\\\\ 2, &otherwise\end{cases}$$:
+Pdf of X is defined as the following
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+$$f(x) = \begin{cases} x for x \le x \e 1, &1-x \le 1 x le 2, &otherwise\end{cases}$$:
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+
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+$E[X] = \int_0^1x \cdot x dx + \int_1^2 x(2-x)dx = \frac{1}{3} + \frac{2}{3} = 1$
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