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(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 enter image description here

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

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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 at 15:53
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1 Answer

$$ \text{E}[\exp(5V)] = \int_1^3 \exp(5v) \, \underbrace{f(v)}_{\text{density}} \, \mathrm{d}v $$ 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.

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thanks alot for your help, I thought that the answer is incorrect but the expected value in "uniformly distributed over the interval[a,b]"=(a+b)/2 I cannot understand your answer what is the diff. between E[v] and exp(v) and I didn't see the law "f(x)=1/b−a" before if you can describe it to me I'll be thankful –  user36894 Jan 5 at 18:15
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I am sorry but there is not much I can help you with... 1) E(V) denotes the expected value of V. exp(V) denotes the exponential of V. 2) The uniform density over (a,b) is constant over (a,b); as it integrates to 1, it has to equal 1/(b-a). –  ocram Jan 5 at 19:01
    
I understood it , you are so helpful thanks a lot –  user36894 Jan 7 at 9:33
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