Hi I am to prove this: E(X) =int_0^infinity [1 − F (x)]dx. We are to work from left to right. That is, start from the definition of the expectation.

1. Start from the definition of the expectation:
E(X) = \int_0^\infty x f(x) \, dx

2. Use the integration by parts formula, where \( u = x \) and \( dv = f(x) \, dx \):
\int u \, dv = uv - \int v \, du

3. Compute \( du \) and \( v \):
du = dx
v = \int f(x) \, dx = F(x)
(Where \( F(x) \) is the cumulative distribution function)

4. Apply integration by parts:
\int_0^\infty x f(x) \, dx = \Big[ x F(x) \Big]_0^\infty - \int_0^\infty F(x) \, dx

5. Evaluate the boundary term:
\Big[ x F(x) \Big]_0^\infty = \lim_{x \to \infty} x F(x) - 0 \cdot F(0) = 0 \cdot 1 - 0 = 0
(Since \( F(\infty) = 1 \) and \( F(0) = 0 \))

6. Hence, we have:
E(X) = 0 - \int_0^\infty F(x) \, dx = -\int_0^\infty F(x) \, dx

7. Note that \( 1 - F(x) \) is valid for any \( x \):
E(X) = \int_0^\infty [1 - F(x)] \, dx

Answer:
E(X) = \int_0^\infty [1 - F(x)] dx