For the definite integral

$\int _{-1}^1\:\left(\frac{27}{x^4}-3\right)dx$

Khan Academy says that the answer is -24, but my TI-84, WolframAlpha, Symbolab, Desmos, pretty much all software says it's undefined. I'm guessing that Khan Academy is correct - the steps make sense and it's written by a human.

KA says that the steps are:

1. Power rule:


$$= \left(9x^{-3}-3x)\right|_{-1}^1$$

2. "Plug in limits of integration":

$$[-9*1^{-3}-3*1]-[-9*(-1)^{-3}-3*(-1)] = -12-12 = -24$$

3. Answer:

$$\int _{-1}^1\:\left(\frac{27}{x^4}-3\right)dx = -24$$

My questions are, is my assumption that KA is correct valid, and what might cause everything else to be wrong?

  • 4
    $\begingroup$ Let $f(x)=27x^{-4}-3$. Is $f(x)$ integrable across the entire interval of integration? $\endgroup$ – Andrew Chin Sep 7 '19 at 23:09
  • $\begingroup$ Look at the graph of that function in Desmos and you'll see what Sal missed. $\endgroup$ – Matthew Daly Sep 7 '19 at 23:13
  • 1
    $\begingroup$ This is why you should read your theorems carefully, especially their conditions, and if ever unsure, check if the integral makes sense at least graphically. $\endgroup$ – Simply Beautiful Art Sep 7 '19 at 23:15
  • $\begingroup$ Not to mention, $27x^{-4} > 3$ for most of that interval. There's no way that can give you a negative number as an answer. $\endgroup$ – Ninad Munshi Sep 7 '19 at 23:17
  • $\begingroup$ That makes sense. I suppose I was over (or under?) thinking it. Thanks! $\endgroup$ – tanner Sep 7 '19 at 23:23

The answer is undefined.

The problem with the solution you quote is that the integrand increases without limit as you approach $x=0$. You cannot apply this method when the interval contains such a point.

You should always check this before substituting limits in this way.

If you look it up you will find lots of information about such "improper integrals".


First, a plot of your integrand.

graph of the integrand

Notice that this is a positive integrand for the entire interval of integration. If the integral exists, its value is positive. The graph also makes clear that this is an improper integral -- the integrand is not continuous at $x = 0$. We must rewrite it as $$ \lim_{r_1 \rightarrow 0^-} \int_{-1}^{r_1} \; \frac{27}{x^4} - 3 \,\mathrm{d}x + \lim_{\ell_2 \rightarrow 0^+} \int_{\ell_2}^{1} \; \frac{27}{x^4} - 3 \,\mathrm{d}x \text{.} $$ Applying the power rule as you did, we end up with $$ \lim_{r_1 \rightarrow 0^-} \left( 3r_1 - \frac{9}{r_1^3} - 12 \right) + \lim_{\ell_2 \rightarrow 0^+} \left( 3\ell_2 + \frac{9}{\ell_2^3} - 12 \right) \text{.} $$ Of course, neither limit exists ...


What is assumed in Step 1 is the Fundamental Theorem of Calculus, which consists of two closely related statements:

Theorem. (FToC) Suppose that $f$ is a continuous function on $[a, b]$. Then

  1. $f$ always have an antiderivative on $[a, b]$.
  2. For any antiderivative $F$ of $f$, we have $$ \int_{a}^{b} f(x) \, \mathrm{d}x = F(b) - F(a).$$

Now let us return to the problem. The function $f(x) = 27x^{-4} - 3$ is not continuous at $x = 0$, and so, FToC cannot be applied directly to the integral in question. This is one issue with this step.

But most of all, it is not clear whether the integral of $f$ even makes sense or not. Since $f$ is not even bounded on $[-1, 1]$, the integral cannot be interpreted in Riemann-integral sense, rendering it undefined in that sense. One may regard it either as improper Riemann-integral (which is the limit of Riemann integrals) or as Lebesgue integral. In such case, notice that

$$ \begin{gathered} \int_{-1}^{-\delta} f(x) \, \mathrm{d}x = \frac{9}{\delta^3} - 12 + 3\delta\\ \int_{\epsilon}^{1} f(x) \, \mathrm{d}x = \frac{9}{\epsilon^3} - 12 + 3\epsilon \end{gathered}$$

holds for all $\delta, \epsilon > 0$. (This is computed by FToC, which is now applicable since $f$ is continuous both on $[\epsilon, 1]$ and on $[-1, -\delta]$.) Then letting $\delta \to 0^+$ and $\epsilon \to 0^+$ simultaneously, this diverges to $+\infty$, and so,

$$ \int_{-1}^{1} \left( \frac{27}{x^4} - 3 \right) \, \mathrm{d}x = \lim_{\delta, \epsilon \to 0^+} \bigg( \int_{-1}^{-\delta} f(x) \, \mathrm{d}x + \int_{\epsilon}^{1} f(x) \, \mathrm{d}x \bigg). = +\infty, $$

either as improper Riemann integral or as Lebesgue integral.


More than similar to other answers, consider the two integrals (where $\epsilon >0$) $$I_1=\int_{-1}^{-\epsilon} \left( \frac{27}{x^4} - 3 \right) \,dx=-9+3 \epsilon+\frac{9}{\epsilon ^3}$$ $$I_2=\int^{1}_{\epsilon} \left( \frac{27}{x^4} - 3 \right) \,dx=-12+3 \epsilon+\frac{9}{\epsilon ^3}$$ making $$I=\int_{-1}^{1} \left( \frac{27}{x^4} - 3 \right) \,dx=\lim_{ \epsilon \to 0} (I_1+I_2)=\lim_{ \epsilon \to 0}\left(-21+6 \epsilon+\frac{18}{\epsilon ^3} \right)=+\infty$$


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