What is the difference between improper and proper integral

If this is for your high school calculus class, not much. An improper integral is basically an integral that cannot "exactly" be calculated. For instance:

∞
∫dx/x²
1

As opposed to the proper integral:

2
∫dx/x²
1

So what you do with an improper integral (assuming it DOES have a value) is that you find the antiderivative then take the limit as it goes to whatever the value is that is causing it to be improper. So for the above example:


∞
∫dx/x² = lim(x --> ∞) {-1/x - -1/1} = 1
1


Edit:

There are cases where you have to be VERY careful. Example:

_∞
_∫dx/x²
-∞

If you did what I did above, then you would say that this integral was 0 (since -1/-∞ = 0 and -1/∞ = 0), however this is INCORRECT because there is a discontinuity in this integral @ x = 0, so the ACTUAL integral should be:

_0_____∞
_∫1/x² + ∫1/x²
-∞_____0

Now you see that -1/0's limit does NOT exist and thus the improper integral does NOT have a value (i.e. the limit does NOT exist)

Edit:

It can get worse even still, though, because while the above integral definitely does NOT have a value, the following integral actually DOES:

_∞
_∫dx/x³ = 0
-∞

...but you can say it's zero based on symmetry arguments...I don't think you'll see integrals like that though.

proper ---> finite domain , bounded function , function continuous almost everywhere

if not ----> improper