Improper Integrals

If we had no intuition regarding what the definite integral should be (the area under the curve), we might very well move on not realizing we have made a severe miscalculation. However, we do have some intuition, and we call tell by the looking at the graph of f(x) = 1/x2 that there is no conceivable way the area bound x = −1 and x = 1 could be negative, let alone attain the value −2 (see Figure 1). But what have we done wrong? This example is subtly different than the previous examples we have seen in that there is a discontinuity at x = 0. It turns out that this is exactly the problem. Our next question is how we might approach rectifying this problem. We notice that for any interval wholly to the left of x = 0 (say, x = −1 to x = a where a < 0), we can apply the standard formula, and the same applies for any interval wholly to the right of x = 0 as well (x = b to x = 1 where b > 0). We are allowed to break integrals into parts like this, so we will consider one integral on the right and one of the left. But how can we manipulate the left and right integrals to capture the whole area we are looking for? In fact, the answer is perhaps the simplest tool we have available to us: we simply