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I saw this integral in my textbook. Is this incorrect? Shouldn't it be $-\lambda e^{-\lambda s}$ since the integral of $e^{-\lambda s}$ is $-\frac{1}{\lambda}e^{-\lambda s}$

\begin{align*} f_S(s) &=\lambda^2\int_0^se^{-\lambda s}\,dt\\ &=\lambda^2se^{-\lambda s} \end{align*}

Image.

Em.
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A user
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    You are integrating with respect to t, not s. Where's the t in the integral? – Graham Kemp Mar 05 '16 at 02:43
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    @GrahamKemp yeah as I stated in my comment below, I was thrown off by the fact that they moved $\lambda^{2}$ outside of the integral but left the rest inside. Thought maybe the dt was a typo and was supposed to read ds. In any case, thanks for your answer. – A user Mar 05 '16 at 03:02

2 Answers2

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The integral is with respect to $t$, not $s$.   So $\mathsf e^{-\lambda s}$ is relatively constant to $t$.

$$\begin{align} f(s) & = \lambda^2\int_0^s \mathsf e^{-\lambda s}\operatorname d t \\[1ex] & = \lambda^2 ~\mathsf e^{-\lambda s}\int_0^s \operatorname d t \\[1ex] & = \lambda^2 ~\mathsf e^{-\lambda s}~(s-0)\\[1ex] & = \lambda^2 ~s ~\mathsf e^{-\lambda s} \end{align}$$

Graham Kemp
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$$\lambda^2\int_0^s e^{-\lambda s}\,dt = \lambda^2e^{-\lambda s}\int_0^s\,dt = \lambda^2e^{-\lambda s}\big[t\big] \bigg|_0^s = \lambda^2e^{-\lambda s}s = \lambda^2 se^{-\lambda s}.$$

Em.
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