# Are definite integrals always tied to the 'area under the curve'?

Our teacher asked us to solve some integrals and when I asked if it we are solving for the net area or signed area, they said that they are not asking for the area.

This got me thinking. Consider the integral $$\int_{-1}^{1}\frac{1}{x^{2}}\,dx.$$

We see that $$\int\frac{1}{x^{2}}\,dx = -\frac{1}{x} + C.$$ Then, $$-\frac{1}{x}\bigg|_{-1}^{1} = -2.$$ I know that $$\lim_{x \to 0}\frac{1}{x^{2}} = +\infty.$$ This means that the integral should be divergent. Without considering areas, is this value meaningful?

Edit: The 'area under the curve' means either the net or total area.

• It is because questions like this deserve careful answers that there is significant development of the theory of integration, and that careful definitions are given of the limiting processes involved. See, for example en.wikipedia.org/wiki/Riemann_integral for the Riemann Integral. The general theory of integration takes into account signed areas (and volumes etc). The integral of a positive function cannot be negative (it may be undefined). Integration is not just about "area" and some definitions will allow you to integrate functions which have no intuitive notion of area. May 11 at 15:17
• By positive function, you mean functions that are positive in the entire domain? May 11 at 15:21
• @MarkBennet "some definitions will allow you to integrate functions which have no intuitive notion of area": Interesting comment. The wikipedia article that you cited referred to alternative definitions of the integral, besides the Reimann integral. Is this what you are referring to? May 11 at 15:23
• $\frac 1{x^2}$ is everywhere non-negative, and undefined at $x=0$. Any method which does integrate this function (with an appropriate fix at zero) will give a positive answer. May 11 at 15:27
• @user2661923 So one can integrate the function which is equal to $1$ at rational values and $0$ at irrational ones over the interval $[0,1]$ for example. The value comes out at zero - but if you have a natural intuition for the area beneath a function which is nowhere continuous you are doing rather better than me. May 11 at 15:30

1. You seem to be applying the fundamental theorem of calculus to evaluate $$\int_{-1}^1 (1/x^2)\,dx$$. But the fundamental theorem of calculus only applies to continuous functions, and $$1/x^2$$ is discontinuous at 0. So your calculation that the value of the integral is -2 is incorrect; you cannot use the fundamental theorem of calculus to evaluate this integral. (Of course, $$1/x^2$$ is always positive, so the fact that your answer is negative should be a tipoff that something is wrong.)
2. You are right that $$\lim_{x \to 0} 1/x^2 = \infty$$. But that doesn't tell you that the integral diverges, it just tells you that it is an improper integral, which means that it should be evaluated like this: $$\int_{-1}^1 \frac{1}{x^2}\,dx = \int_{-1}^0 \frac{1}{x^2}\,dx + \int_0^1 \frac{1}{x^2}\,dx = \lim_{a \to 0^-} \int_{-1}^a \frac{1}{x^2}\,dx + \lim_{b \to 0^+} \int_b^1 \frac{1}{x^2}\,dx.$$ Whether the integral converges or diverges is determined by whether or not the limits on the right-hand side above are defined.
• Further to point $2$, in some cases a function that diverges at $0$ has a unique well-defined integral on $[-1,\,1]$ no matter how you define integrals, e.g. $|x|^{-1/2}$.