this diverges. we only have a problem at x = 1, so since that is not between our limits of integration we dont have to split the integral up.

So the integral diverges

looking at the graph and seeing that is has a limit means nothing. look at the graph of , it's limit is 0 as x goes to , but its improper integral diverges

this converges.2. int {e^2x} dx from -infinity to 0

i put it diverges because the limit does not exist. i got [(e^2x)/2] 0 to N. then i put (e^0 / 2) - (e^2N / 2)

and then i looked at the graph of the integral.

Thus the integral converges

remember, when we raise something to a negative exponent, it's the same as putting 1 over the something to the positive exponent. So