$\"\"$

The statement is $\"\"$ .

Continuity Implies Integrability Theorem:

If a function $\"\"$ is continuous on the closed interval $\"\"$ , then $\"\"$ is integrable on $\"\"$ .

$\"\"$ exists.

Consider the function $\"\"$ .

Graph of the function $\"\"$ :

$\"\"$

Observe the graph:

The function is not continuous on the interval $\"\"$ .

$\"\"$

Each of these integral is infinite.

The function is not integrable on $\"\"$ .

$\"\"$ has a non removable discontinuity at $\"\"$ .

$\"\"$

$\"\"$ has a non removable discontinuity at $\"\"$ .