Are definite integrals always tied to the ‘area under the curve’

areacalculusdefinite integrals

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.

Best Answer

Definite integrals (of the kind studied in calculus classes--that is, Riemann integrals) are always tied to Riemann sums, and Riemann sums can be thought of as representing (signed) areas. So while your teacher may not have been explicitly asking for an area, the answer could be interpreted as a (signed) area.

There are two mistakes in what you have written:

  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.
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