Question

Let \(X\) have the pdf \(f(x)=(x+2) / 18,-2

Short answer

While the exact solutions to these integrals do depend on your ability to solve them correctly, it will be a numerical value for each of the expected values \(E(X)\), \(E[(X+2)^3]\), and \(E[6X-2(X+2)^3]\).

Step-by-step solution

Find E(X)

Firstly, you need to find the expected value \(E(X)\). This involves integrating the product of \(X\) and the pdf, \(f(x)\), over the range \(x\) (-2, 4). Query:\[\int_{-2}^{4} x \cdot f(x) dx = \int_{-2}^{4} x \cdot \frac{x+2}{18} dx\]

Find E[(X+2)^3]

Next, the task is to find the expected value \(E[(X+2)^3]\). This involves integrating the product of \((X+2)^3\) and the pdf, \(f(x)\), over the range \(x\) (-2, 4). Query:\[\int_{-2}^{4} (x+2)^3 \cdot f(x) dx = \int_{-2}^{4} (x+2)^3 \cdot \frac{x+2}{18} dx\]

Find E[6X-2(X+2)^3]

Finally, you need to find the expected value \(E[6X-2(X+2)^3]\). This involves integrating the product of \([6X-2(X+2)^3]\) and the pdf, \(f(x)\), over the range \(x\) (-2, 4). Query:\[\int_{-2}^{4} [6X-2(X+2)^3] \cdot f(x) dx = \int_{-2}^{4} [6X-2(X+2)^3] \cdot \frac{x+2}{18} dx\]