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**Jhevon** we have improper integrals when the limits of our integrals are not nice. and by not nice, i mean it makes no sense to plug them into our original integrand, or sometimes, even in our integral for that matter. something like $\displaystyle \int_0^\infty \sin x~dx$ is an improper integral. why? because it makes no sense to plug in $\displaystyle \infty$ into something. $\displaystyle \infty$ is not a real number. (that's why whenever we talk about $\displaystyle \infty$ we talk about the limit as we approach $\displaystyle \infty$ and not $\displaystyle \infty$ itself). an integral like $\displaystyle \int_0^1 \frac 1x~dx$ is also improper. why? because $\displaystyle \frac 1x$ is not defined at 0, how can we be evaluating an integral there. that's crazy talk!

the method of improper integrals allow us to deal with such problems. how? well, the way we are familiar with. how do we usually try to evaluate something at a point we can't plug in? we use limits!

so, here's how it works. we replace the bad limit point with a variable. i think "N" is the usual one, but it can be whatever. and then we use limits to evaluate the integral at that point.

example, going back to $\displaystyle \int_0^\infty \sin x~dx$. we would write this as $\displaystyle \lim_{N \to \infty} \int_0^N \sin x~dx$

now we evaluate the integral using N in place of $\displaystyle \infty$, then we plug in the limits and take the limit as $\displaystyle N \to \infty$.

now, not all improper integrals exist. we can only get an answer if we get definite answer for the limit of the integral. otherwise, we say the improper integral "diverges" or "does not exist" whichever's more appropriate.

so for you problem:

$\displaystyle \int_{-1}^2 \frac 3x~dx = \int_{-1}^0 \frac 3x~dx + \int_0^2 \frac 3x~dx$

$\displaystyle = \lim_{N \to 0} \int_{-1}^N \frac 3x~dx + \lim_{A \to 0} \int_A^2 \frac 3x~dx$

now evaluate both integrals and take the limits.