Barrons AP Calculus - David Bock

Improper integrals of class (1), where the interval is not bounded, are handled as limits:

where f is continuous on [a,b]. If the limit on the right exists, the improper integral on the left is said
to converge to this limit; if the limit on the right fails to exist, we say that the improper integral
diverges (or is meaningless).
The evaluation of improper integrals of class (1) is illustrated in Examples 17–23.
EXAMPLE 17
Find
SOLUTION: The given integral thus converges to 1. In
Figure N7–22 we interpret as the area above the x-axis, under the curve of y = 3, and bounded
at the left by the vertical line x = 1.

FIGURE N7–22

BC ONLY

EXAMPLE 18

It can be proved that converges if p > 1 but diverges if p 1. Figure N7–23 gives a

geometric interpretation in terms of area of Only the first-quadrant area under
bounded at the left by x = 1 exists. Note that