Let $X$ be uniformly distributed in the interval $[0, 2].$

Question

(a) Compute $E(e^X)$

(b) Compute $E(\sin(X))$

I am confused that how to compute the interval since it did not give the formula

Answer 1

What you want to do is the following:

(a) $\displaystyle\int_0^2 \dfrac{1}{2}e^x dx$, which is $\dfrac{e^2-1}{2}$, and

(b) $\displaystyle\int_0^2 \dfrac{1}{2}\sin x dx$, which is $\dfrac{1-\cos2}{2}$.

Answer 2

Let $\phi(x)$ be the density of $X$.

Since $X \sim U[0,2]$, it follows that $\phi(x) = 1/2$ on the interval $[0, 2]$ and $\phi(x) = 0$ outside of the interval $[0, 2]$.

Therefore, $\lower{7.5ex}{\begin{align*} \mathbb{E}[f(X)] &= \int_{-\infty}^{\infty} \phi(x) f(x) dx \\[1ex]&= \int_{-\infty}^{0} 0 \cdot f(x) dx + \int_0^2 \frac{1}{2} f(x) dx + \int_2^\infty 0 \cdot f(x) dx \\[1ex]&= \int_0^2 \frac{1}{2} f(x) dx. \end{align*}}$

In your case, you have two different $f$.

Substitute each one into the above and compute the resulting integral.