The length $X$ of of a call follows the exponential distribution with mean $2$ minutes. In dollars, the cost of of a call of $x$ minutes is $3x^2-6x+2$. Find the expected cost of a call?
The addition of the $3x^2-6x+2$ is what is throwing me off, as I know to find $E(x)$ of exponential distributions is just $\dfrac{1}{\lambda}$, meaning the rate $\lambda$ is equal to $\dfrac{1}{2}$.
Any help on solving this is helpful!
To find the expected value of something, you multiply each of its values by the probability of those values, and add them all up, which in this case means you integrate. $$E(3X^2 - 6X + 2) = \int_0^{\infty} (3x^2 - 6x + 2)f(x)dx$$ where f(x) is the pdf of the exponential distribution, so $$E(3X^2 - 6X + 2) = \int_0^{\infty} (3x^2 - 6x + 2)\frac{1}{2}e^{-x/2}dx.$$ Note that "the probability of those values" here means $f(x)dx$ = $e^{-x/2}dx$ which becomes a probability of a cost in some interval when we integrate. It's not $f(x) = e^{-x/2}$ itself. But you don't have to integrate all of of that because $$E(3X^2 - 6X + 2) = E(3X^2) - E(6X) + 2$$ $$= 3E(X^2) - 6E(X) + 2.$$ You already know $E(X) = 2$, so you just need $E(X^2)$, so find that like above. $$E(X^2) = \int_0^{\infty} x^2\frac{1}{2}e^{-x/2}dx$$