Calculate the improper integral: integral from 0 to ²

Calculate the improper integral: $$ \int_0^2 \frac{dx}{\sqrt[3]{(1-x)^2}} $$

Step 1: Simplify the integrand

$$\frac{1}{\sqrt[3]{(1-x)^2}} = \frac{1}{[(1-x)^2]^{1/3}} = \frac{1}{(1-x)^{2/3}} = (1-x)^{-2/3}$$

Note: $(1-x)^2 = |1-x|^2$, so:

$$[(1-x)^2]^{1/3} = |1-x|^{2/3}$$

Step 2: Identify the singularity

The integrand has a singularity at $x = 1$, where the denominator $(1-x)^{2/3} \to 0$ and the function goes to infinity.

For the interval $[0, 2]$:

• For $x \in [0, 1)$: $1-x > 0$, so $|1-x|^{2/3} = (1-x)^{2/3}$
• For $x \in (1, 2]$: $1-x < 0$, so $|1-x|^{2/3} = (x-1)^{2/3}$

We must split the integral at $x = 1$:

$$\int_0^2 \frac{dx}{\sqrt[3]{(1-x)^2}} = \int_0^1 \frac{dx}{(1-x)^{2/3}} + \int_1^2 \frac{dx}{(x-1)^{2/3}}$$

Step 3: Check convergence

Both integrals are of the form $\int \frac{dx}{u^\alpha}$ with $\alpha = \frac{2}{3}$.

Convergence criterion: The integral $\int_a^b \frac{dx}{(b-x)^\alpha}$ converges if and only if $\alpha < 1$.

Since $\frac{2}{3} < 1$, both integrals converge! ✓

Step 4: Evaluate the first integral $\int_0^1 \frac{dx}{(1-x)^{2/3}}$

$$\int_0^1 (1-x)^{-2/3} \, dx = \lim_{\varepsilon \to 0^+} \int_0^{1-\varepsilon} (1-x)^{-2/3} \, dx$$

Using the power rule:

$$\int (1-x)^{-2/3} \, dx = \frac{(1-x)^{1/3}}{-1 \cdot \frac{1}{3}} = -3(1-x)^{1/3} + C$$

Verify:

$$\frac{d}{dx}[-3(1-x)^{1/3}] = -3 \cdot \frac{1}{3}(1-x)^{-2/3} \cdot (-1) = (1-x)^{-2/3}$$

Correct! ✓

Evaluating:

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

$$= -3(1-(1-\varepsilon))^{1/3} - (-3(1-0)^{1/3}) = -3\varepsilon^{1/3} + 3$$

Taking the limit:

$$\lim_{\varepsilon \to 0^+} (-3\varepsilon^{1/3} + 3) = 0 + 3 = 3$$

Step 5: Evaluate the second integral $\int_1^2 \frac{dx}{(x-1)^{2/3}}$

$$\int_1^2 (x-1)^{-2/3} \, dx = \lim_{\varepsilon \to 0^+} \int_{1+\varepsilon}^2 (x-1)^{-2/3} \, dx$$

Using the power rule:

$$\int (x-1)^{-2/3} \, dx = \frac{(x-1)^{1/3}}{\frac{1}{3}} = 3(x-1)^{1/3} + C$$

Evaluating:

$$\int_{1+\varepsilon}^2 (x-1)^{-2/3} \, dx = [3(x-1)^{1/3}]_{1+\varepsilon}^2$$

$$= 3(2-1)^{1/3} - 3((1+\varepsilon)-1)^{1/3} = 3(1) - 3\varepsilon^{1/3}$$

Taking the limit:

$$\lim_{\varepsilon \to 0^+} (3 - 3\varepsilon^{1/3}) = 3 - 0 = 3$$

Step 6: Combine the results

$$\int_0^2 \frac{dx}{\sqrt[3]{(1-x)^2}} = 3 + 3 = 6$$

Final Answer

$$\boxed{\int_0^2 \frac{dx}{\sqrt[3]{(1-x)^2}} = 6}$$

The integral converges to 6.

Verification using the general formula

From the general result, for $y = 3$:

$$\int_0^2 \frac{dx}{\sqrt[y]{(1-x)^2}} = \frac{2y}{y-2} = \frac{2(3)}{3-2} = \frac{6}{1} = 6$$

This confirms our answer! ✓

Alternative verification by direct computation

Let's verify by computing directly without limits:

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

$$\int_1^2 (x-1)^{-2/3} \, dx = [3(x-1)^{1/3}]_1^2 = 3(1) - 3(0) = 3 - 0 = 3$$

Total: $3 + 3 = 6$ ✓

Geometric interpretation

The function $f(x) = \frac{1}{|1-x|^{2/3}}$ has these properties:

• Singularity at $x = 1$ (goes to $+\infty$)
• Always positive on $[0, 2]$
• Symmetric about $x = 1$: $f(1-a) = f(1+a)$
• At $x = 0$: $f(0) = \frac{1}{1^{2/3}} = 1$
• At $x = 2$: $f(2) = \frac{1}{1^{2/3}} = 1$

The area under the curve from 0 to 2 is finite (equals 6) despite the infinite spike at $x = 1$.

Why the integral converges

The exponent $\alpha = \frac{2}{3}$ is less than the critical value of 1, so:

• Near $x = 1^-$: $\frac{1}{(1-x)^{2/3}}$ grows to infinity, but not "too fast"
• Near $x = 1^+$: $\frac{1}{(x-1)^{2/3}}$ grows to infinity, but not "too fast"
• The area remains finite because $\frac{2}{3} < 1$

If the exponent were $\geq 1$, the integral would diverge.

Comparison with other exponents

Case 1: $\int_0^2 \frac{dx}{\sqrt[4]{(1-x)^2}} = \int_0^2 \frac{dx}{|1-x|^{1/2}}$ (exponent $\frac{1}{2} < 1$)

$$= \frac{2(4)}{4-2} = 4 \quad \text{(converges)}$$

Case 2: $\int_0^2 \frac{dx}{\sqrt[3]{(1-x)^2}} = \int_0^2 \frac{dx}{|1-x|^{2/3}}$ (exponent $\frac{2}{3} < 1$)

$$= \frac{2(3)}{3-2} = 6 \quad \text{(converges)}$$

Case 3: $\int_0^2 \frac{dx}{\sqrt[2]{(1-x)^2}} = \int_0^2 \frac{dx}{|1-x|}$ (exponent $1$)

$$= +\infty \quad \text{(diverges - critical case)}$$

Case 4: $\int_0^2 \frac{dx}{(1-x)^2}$ (exponent $2 > 1$)

$$= +\infty \quad \text{(diverges)}$$

As the exponent increases toward 1, the integral value increases. At $\alpha = 1$, it diverges.

Summary

1. The integrand is $\frac{1}{|1-x|^{2/3}}$ with singularity at $x = 1$
2. We split: $\int_0^2 = \int_0^1 + \int_1^2$
3. The exponent $\alpha = \frac{2}{3} < 1$ ensures convergence
4. Each integral evaluates to 3: \begin{itemize}
5. $\int_0^1 \frac{dx}{(1-x)^{2/3}} = 3$
6. $\int_1^2 \frac{dx}{(x-1)^{2/3}} = 3$ \end{itemize}
7. Total: $\int_0^2 \frac{dx}{\sqrt[3]{(1-x)^2}} = 6$
8. This matches the general formula: $\frac{2y}{y-2} = \frac{6}{1} = 6$ for $y = 3$

Key lessons:

• Cube roots (and higher roots) often lead to convergent integrals
• The exponent $\frac{2}{3}$ is "weak enough" for convergence
• Symmetry about the singularity means both halves contribute equally
• The critical exponent $\alpha = 1$ separates convergence from divergence
• Fractional exponents less than 1 allow integrals to converge despite singularities
• This is a concrete example where an "infinite spike" has finite area underneath