Leeward 206 {8.13} Integral at infinity

[karush]

$\tiny\text{Leeward 206 {8.13} Integral at infinity}$
$$I=\int_{0} ^{\infty} e^{-ax} \,dx \ a>0 =
\\
\begin{align}\displaystyle
u& = -ax &
du&=-a \ d{x}
\end{align} \\
\text{then} \\
I=-\frac{1}{a}\int_{0} ^{\infty} e^{x} \,dx
=-\dfrac{\mathrm{e}^{-ax}}{a}+C \\
\text{hopefully, wasn't sure about the + C}$$
$\tiny\text{ Surf the Nations math study group}$

 

[MarkFL]

When you make the substitution you need to change your limits in accordance with the substitution. Let's begin with:

\(\displaystyle I=\lim_{t\to\infty}\left(\int_0^t e^{-ax}\,dx\right)\) where $0<a$

Now, if we use the substitution:

\(\displaystyle u=-ax\,\therefore\,du=-a\,dx\implies dx=-\frac{1}{a}\,du\)

and use the rule:

\(\displaystyle \int_a^b f(x)\,dx=-\int_b^a f(x)\,dx\)

We obtain:

\(\displaystyle I=\frac{1}{a}\lim_{t\to\infty}\left(\int_{-t}^0 e^{u}\,du\right)=\frac{1}{a}\lim_{t\to\infty}\left(\left[e^u\right]_{-t}^0\right)=\frac{1}{a}\lim_{t\to\infty}\left(1-e^{-t}\right)=\frac{1}{a}\)

 

[karush]

$\text{Why did you introduce t ?}$

 

[MarkFL]

$\text{Why did you introduce t ?}$

That's how I was taught to handle improper integrals. :)

 

[Prove It]

$\text{Why did you introduce t ?}$

The reason Mark was taught that is how you handle improper integrals is because a definite integral requires FINITE values for the terminals of the integral. So by applying a finite terminal (in this case, "t") you can then evaluate the improper integral by seeing what happens as that finite value is increased without bound, in other words, what is the limiting value as $\displaystyle \begin{align*} t \to \infty \end{align*}$...