Integral of exponential functions

Question

On this page, there are two integrals of exponential functions.

First,

$$\int_{0}^{\infty} c\cdot N_{0}e^{-\lambda t}dt=c\cdot \frac{N_{0}}{\lambda}$$

How does one get this result? I got $\int_{0}^{\infty} c\cdot N_{0}e^{-\lambda t}dt=\frac{c\cdot N_{0}\cdot e^{-\lambda t}}{-\lambda}$ instead.

Second,

$$\int_{0}^{\infty} \lambda t\cdot e^{-\lambda t}dt=\frac{1}{\lambda}$$

I thought we'd solve this using integration by parts, and then I got

$$\int_{0}^{\infty} \lambda t\cdot e^{-\lambda t}dt=\lambda t\cdot \frac{e^{-\lambda t}}{-\lambda}-\int_{0}^{\infty}\frac{e^{-\lambda t}}{-\lambda}\cdot \lambda dt=-t\cdot e^{-\lambda t}-\frac{e^{-\lambda t}}{\lambda}=-e^{-\lambda t}(t+\frac{1}{\lambda})$$

Again, I am puzzled that how one gets the $\frac{1}{\lambda}$ result.

I may be missing something here, and some explanation would be appreciated.

Answer 1

$$\begin{align} \int_0^{\infty}cN_0e^{-\lambda t}dt&=\left. -cN_0\frac{e^{-\lambda t}}{\lambda}\right|_0^{\infty}\\\\ &=0-\left(-N_0c\frac{e^{-\lambda 0}}{\lambda}\right)\\\\ &=\frac{cN_0}{\lambda} \end{align}$$

---

$$\begin{align} \int_0^{\infty}\lambda te^{-\lambda t}dt&= \left. -te^{-\lambda t}\right|_0^{\infty}+\int_0^{\infty}e^{-\lambda t}dt\\\\ &=\frac{1}{\lambda} \end{align}$$

Answer 2

Put $\lambda t = u$, then

$$\int_{0}^{\infty} c N_{0}e^{-\lambda t}dt = \frac{c N_{0}}{\lambda}\int_{0}^{\infty}e^{-u} \, du.$$

Same substitution for the second one, get $$\frac{1}{\lambda}\int_{0}^{\infty} u e^{-u} \, du.$$

These are both easy to integrate. You should be able to continue.