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Consider the density function $f(x)$ = ${e^x}\over{e - 1}$, with $0 < x < 1$. Find:

$a)$ $F(.4)$

$b)$ $P(.4< x <.6)$

For $(b)$ I computed $\int_{.4}^{.6}$$e^x\over{e-1}$ and obtained .1922. For $(a)$ I computed $\int_{0}^{.4}$$e^.4\over{e-1}$ and got .8682. I think I did something incorrectly for part $(a)$ however, since I would expect the answer to be less than 1-.1922 where .1922 is my result from $(b)$.

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    $\int_{0}^{0.4} \frac{e^{x}}{e-1}\operatorname{d}\!x \neq \frac{e^{0.4}}{e-1}\approx 0.8682$ – 2017-02-01
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    Oops, accidentally put the upper limit to be 1 – 2017-02-01

1 Answers 1

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Note that in taking the integral you can move the denominator to the outside as it is constant and since the integral of $\int_{x}e^{x} dx = e^{x}$ we have

$\int_{0}^{.4} f(x) dx = \frac{e^{.4} - e^{0}}{e - 1} \approx \frac{.49}{1.72} \approx .28$

$\int_{.4}^{.6} f(x) dx = \frac{e^{.6} - e^{.4}}{e - 1} \approx \frac{.33}{1.72} \approx .19$

which is all nice and consistent.

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    you should write in math mode – 2017-02-01