LCC 8.8.11 Infinite Intervals of Integration

[karush]

$\tiny\text{LCC 206 8.8.11 Infinite Intervals of Integration}$
$$\displaystyle
I=\int_{1}^{\infty} {x}^{-2} \,dx = 1$$
$$I=\left[\frac{1}{x}\right]_1^\infty=\left| 0-1 \right|=1$$

$\text{the only way apparently to get 1 is to use absolute value ?}$

$\tiny\text{from Surf the Nations math study group}$

 

[MarkFL]

I would write:

$$I=\lim_{t\to\infty}\left(\int_1^t x^{-2}\,dx\right)=\lim_{t\to\infty}\left(\left[-\frac{1}{x}\right]_1^t\right)=\lim_{t\to\infty}\left(1-\frac{1}{t}\right)=1$$