Evaluate the integral of 1 / x^(2/3) from x = -1 to x = 1.

Step-by-step solution

1. 1

   Spot the interior singularity. The integrand $x^{-2/3}$ is undefined at $x=0$, which lies inside the interval of integration. Applying the fundamental theorem straight across the gap is invalid; the integral must be treated as improper and split at $0$:

   $\int_{-1}^{1}x^{-2/3}dx=\int_{-1}^{0}x^{-2/3}dx+\int_{0}^{1}x^{-2/3}dx$

2. 2

   Find the antiderivative. By the power rule with $p=-\tfrac23$, so $p+1=\tfrac13$:

   $\int x^{-2/3}dx=\frac{x^{1/3}}{1/3}=3x^{1/3}$

   The real cube root $x^{1/3}$ is defined for negative $x$ too, so this single formula covers both halves.

3. 3

   Evaluate the left half as a limit.

   $\lim_{a\to 0^{-}}\left[3x^{1/3}\right]_{-1}^{a}=\lim_{a\to 0^{-}}\left(3a^{1/3}-3(-1)^{1/3}\right)=0+3=3$

   using $(-1)^{1/3}=-1$. The limit is finite because $x^{1/3}\to 0$, not $\infty$: the antiderivative is continuous even where the integrand is not.

4. 4

   Evaluate the right half the same way.

   $\lim_{b\to 0^{+}}\left[3x^{1/3}\right]_{b}^{1}=\lim_{b\to 0^{+}}\left(3-3b^{1/3}\right)=3$

5. 5

   Add the two convergent halves.

   $\int_{-1}^{1}\frac{1}{x^{2/3}}dx=3+3=6$

   The integral converges even though the integrand is unbounded — the spike at $0$ is simply too narrow to enclose infinite area.

6. 6

   Compare with the general rule. Near $0$, $\int_{0}^{1}x^{-p}dx$ converges exactly when $p<1$. Here $p=\tfrac23<1$, so convergence was predictable; had the exponent been $x^{-4/3}$ (so $p=\tfrac43>1$) both halves would diverge. The symmetry of the integrand also explains why both halves give the same $3$.