Calculate the integral: integral from 0 to 1 dx/(sqrt(x))

Calculate the integral: $\int_0^1 \frac{dx}{\sqrt{x}}$

This is an improper integral because the integrand $\frac{1}{\sqrt{x}}$ has a vertical asymptote at $x = 0$ (it approaches infinity as $x \to 0^+$).

Step 1: Recognize the improper integral

The function $f(x) = \frac{1}{\sqrt{x}} = x^{-1/2}$ is not defined at $x = 0$ and:

$$\lim_{x \to 0^+} \frac{1}{\sqrt{x}} = +\infty$$

Therefore, we must evaluate this as an improper integral:

$$\int_0^1 \frac{dx}{\sqrt{x}} = \lim_{\varepsilon \to 0^+} \int_\varepsilon^1 \frac{dx}{\sqrt{x}}$$

Step 2: Find the antiderivative

We need to find:

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

Using the power rule for integration:

$$\int x^n \, dx = \frac{x^{n+1}}{n+1} + C \quad \text{for } n \neq -1$$

With $n = -\frac{1}{2}$:

$$\int x^{-1/2} \, dx = \frac{x^{-1/2+1}}{-1/2+1} + C = \frac{x^{1/2}}{1/2} + C = 2x^{1/2} + C = 2\sqrt{x} + C$$

Step 3: Evaluate the definite integral

$$\int_\varepsilon^1 \frac{dx}{\sqrt{x}} = \left[2\sqrt{x}\right]_\varepsilon^1 = 2\sqrt{1} - 2\sqrt{\varepsilon} = 2 - 2\sqrt{\varepsilon}$$

Step 4: Take the limit

$$\int_0^1 \frac{dx}{\sqrt{x}} = \lim_{\varepsilon \to 0^+} (2 - 2\sqrt{\varepsilon}) = 2 - 2 \cdot 0 = 2$$

Final Answer

$$\boxed{\int_0^1 \frac{dx}{\sqrt{x}} = 2}$$

Verification

Let's verify this result makes sense:

Check the antiderivative:

$$\frac{d}{dx}[2\sqrt{x}] = 2 \cdot \frac{1}{2\sqrt{x}} = \frac{1}{\sqrt{x}}$$

This confirms our antiderivative is correct. ✓

Convergence check:

The integral converges because near $x = 0$:

$$\frac{1}{\sqrt{x}} = \frac{1}{x^{1/2}}$$

The integral $\int_0^a \frac{dx}{x^p}$ converges if and only if $p < 1$.

Since $p = \frac{1}{2} < 1$, the integral converges. ✓

Geometric interpretation

The integral $\int_0^1 \frac{dx}{\sqrt{x}}$ represents the area under the curve $y = \frac{1}{\sqrt{x}}$ from $x = 0$ to $x = 1$.

Behavior of the function:

• At $x = 0^+$: $y \to +\infty$ (vertical asymptote)
• At $x = 1$: $y = 1$
• The function is decreasing on $(0, 1]$

Despite the function becoming infinite at $x = 0$, the area under the curve is finite (equals 2).

Why the integral converges

Intuitively, although $\frac{1}{\sqrt{x}}$ grows to infinity as $x \to 0^+$, it doesn't grow "too fast" for the integral to diverge.

We can think of the area near $x = 0$ as approximately:

$$\text{Area} \approx \text{height} \times \text{width} \approx \frac{1}{\sqrt{x}} \cdot \Delta x$$

For small $x$, this becomes:

$$\frac{\Delta x}{\sqrt{x}} = \frac{\Delta x}{x^{1/2}} = x^{-1/2} \Delta x$$

The key is that as $x \to 0$, the width $\Delta x$ decreases faster than the height increases, so the total area remains finite.

Comparison with other singularities

Let's compare with similar integrals:

Case 1: $\int_0^1 \frac{dx}{x^{1/3}}$ (weaker singularity)

$$\int_0^1 x^{-1/3} \, dx = \left[\frac{x^{2/3}}{2/3}\right]_0^1 = \frac{3}{2}(1 - 0) = \frac{3}{2}$$

This converges. ✓

Case 2: $\int_0^1 \frac{dx}{x}$ (critical case)

$$\int_0^1 \frac{dx}{x} = \lim_{\varepsilon \to 0^+} [\ln x]_\varepsilon^1 = \lim_{\varepsilon \to 0^+} (0 - \ln\varepsilon) = +\infty$$

This diverges! ✗

Case 3: $\int_0^1 \frac{dx}{x^{3/2}}$ (stronger singularity)

$$\int_0^1 x^{-3/2} \, dx = \left[\frac{x^{-1/2}}{-1/2}\right]_0^1 = -2\left[\frac{1}{\sqrt{x}}\right]_0^1 = -2(1 - \infty) = +\infty$$

This diverges! ✗

General rule:

$$\int_0^1 \frac{dx}{x^p} \begin{cases} \text{converges} & \text{if } p < 1\\ \text{diverges} & \text{if } p \geq 1 \end{cases}$$

Our case has $p = \frac{1}{2} < 1$, so it converges.

Alternative calculation using substitution

We can also solve this using substitution. Let $u = \sqrt{x}$, so $x = u^2$ and $dx = 2u \, du$.

When $x = 0$: $u = 0$

When $x = 1$: $u = 1$

$$\int_0^1 \frac{dx}{\sqrt{x}} = \int_0^1 \frac{2u \, du}{u} = \int_0^1 2 \, du = [2u]_0^1 = 2(1) - 2(0) = 2$$

This gives the same answer and shows that the substitution removes the singularity!

Physical interpretation

This type of integral appears in physics, for example:

Time for a particle to fall from height $h$ to 0:

Under gravity, the velocity is $v = \sqrt{2gh}$, so $\frac{dx}{dt} = \sqrt{2gx}$ (taking $x$ as height).

The time to fall is:

$$t = \int_0^h \frac{dx}{\sqrt{2gx}} = \frac{1}{\sqrt{2g}} \int_0^h \frac{dx}{\sqrt{x}} = \frac{1}{\sqrt{2g}} \cdot 2\sqrt{h} = 2\sqrt{\frac{h}{2g}}$$

The integral converges even though the velocity approaches infinity as the particle nears the ground!

Summary

1. The integral $\int_0^1 \frac{dx}{\sqrt{x}}$ is improper due to a vertical asymptote at $x = 0$
2. We evaluate it as a limit: $\lim_{\varepsilon \to 0^+} \int_\varepsilon^1 \frac{dx}{\sqrt{x}}$
3. The antiderivative is $2\sqrt{x}$
4. The limit converges to $2$
5. The convergence occurs because the exponent $-\frac{1}{2} > -1$
6. Despite the integrand being unbounded, the area under the curve is finite

Key lessons:

• Not all unbounded integrands lead to divergent integrals
• The power rule works for $n \neq -1$, including negative fractional powers
• Improper integrals with singularities at endpoints require limit evaluation
• The critical exponent for convergence at 0 is $p = 1$
• Substitution can sometimes transform an improper integral into a proper one