Find E{1/x} if we are given a density function with continuos random variable

Question

Let X be a continuous random variable with density function

$$f(x) = \begin{cases}\frac{x}{30}(1+3x) & 1 < x < 3 \\0 & \text{otherwise}\end{cases}$$

Find $E\left(\frac1x\right)$

Answer 1

$$E\left[\frac{1}{x}\right]=\int_1^3 \frac{1}{x}\cdot \frac{x}{30}(1+3x)\, dx$$

Now please simplify and do the integration if you want to learn something...

Answer 2

If a random variable $X$ has density function $f(x)$, then the random variable $g(X)$ has mean $$\int_{-\infty}^{\infty} g(x)f(x)\,dx.$$ In your case, $g(x)=\frac{1}{x}$, so the mean of $\frac{1}{X}$ is $$\int_1^3 \frac{1}{x}\frac{x}{30}(1+3x)\,dx.$$ The integration will be easy.