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We are given $f(x)=\frac{1}{b-a}$ for $a<x<b$, and $0$ otherwise. My solution was to integrate $f(x)$ such that: $\frac{1}{b-a}\int_{a}^bdx=\frac{1}{b-a}[x]_{a}^b=\frac{1}{b-a}[b-a]$. I think I'm missing something. Any help would be appreciated, thanks

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  • $\begingroup$ Why have you integrated over the whole domain? $\endgroup$ – soakley Oct 17 '16 at 21:05
  • $\begingroup$ you calculated F(b) $\endgroup$ – oW_ Oct 17 '16 at 21:06
  • $\begingroup$ So I integrate from $a$ to $x$? Can you give me some insight $\endgroup$ – Lanous Oct 17 '16 at 21:08
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    $\begingroup$ Yes, although I'd change your notation slightly (inside the integral - see the proposed answer) to avoid ambiguity. Recall the definition of the CDF. You want to be able to evaluate it at any point in the domain. $\endgroup$ – soakley Oct 17 '16 at 21:11
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You should integrate over $[a, x]$, not $[a, b]$:

$$F(x)=\int_{a}^{x}f(t)dt=\int_a^x\frac{1}{b-a}dt=\frac{1}{b-a}\int_a^xdt=\left[\frac{t}{b-a}\right]_a^x=\frac{x-a}{b-a}$$

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