# how calculate expected value

(Ross [2009], p.162) The current in a semiconductor diode is often measured by the Shockley equation I = I0(e^aV-1) where V is the voltage across the diode; I0 is the reverse current; a is a constant; and I is the resulting diode current. Find E(I ) if a = 5, I0= 10^-6 , and V is uniformly distributed over [1; 3]. Answer

my question is: how "1/2" is calculated ??? E[x]= (a+b)/2 thats mean (1+3)/2 =2 not 1/2 I need help please thanks in advance

-
The solution, although leading to a correct answer, is prima facie nonsense because none of the integrals is syntactically well formed. They are missing a "$dv$" term. Although this might seem picky, it gets to the heart of the question, which comes down to asking why $\frac{1}{2}dv$ (for $1\le v\le 3$) would be the density of a Uniform$[1,3]$ distribution. Without the $dv$ term this is not a density at all, so it is understandable why a careful reader would be confused. – whuber Jan 5 '14 at 15:53

$$\text{E}[\exp(5V)] = \int_1^3 \exp(5v) \, \underbrace{f(v)}_{\text{density}} \, \mathrm{d}v$$ with $$f(v) = 0.5 \quad \text{for} \quad 1 < v < 3$$ as $V$ is uniform over $(1, 3)$.
$X \sim U(a,b)$ has density $f(x) = \frac{1}{b-a}$ for $x \in (a,b)$ and $0$ elsewhere.