Evaluate the improper integral from 0 to 1 of 4/x2 dx.

The integral is $\int_0^1 \frac{4}{x^2} dx$.
We can rewrite $\frac{4}{x^2}$ as $4x^{-2}$.
So the integral becomes:

$$\int_0^1 4x^{-2} dx$$

The integrand $f(x) = \frac{4}{x^2}$ is undefined at $x=0$, which is one of the limits of integration. Therefore, this is an improper integral of Type 2. We evaluate it using a limit:

$$\int_0^1 4x^{-2} dx = \lim_{a \to 0^+} \int_a^1 4x^{-2} dx$$

Using the power rule for integration, $\int x^n dx = \frac{x^{n+1}}{n+1} + C$:

$$\int 4x^{-2} dx = 4 \left( \frac{x^{-2+1}}{-2+1} \right) = 4 \left( \frac{x^{-1}}{-1} \right) = -4x^{-1} = -\frac{4}{x}$$

$$\left[ -\frac{4}{x} \right]_a^1 = \left( -\frac{4}{1} \right) - \left( -\frac{4}{a} \right)$$ $$= -4 + \frac{4}{a}$$

$$\lim_{a \to 0^+} \left( -4 + \frac{4}{a} \right)$$

As $a$ approaches 0 from the positive side, $\frac{4}{a}$ approaches positive infinity.

$$\lim_{a \to 0^+} \left( -4 + \frac{4}{a} \right) = -4 + \infty = \infty$$

Since the limit is $\infty$, the integral diverges.

The final answer is $\boxed{\text{Diverges}}$.
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