Step 1: Understanding the Concept:
We have a definite integral with a fractional power. To simplify it, it is a standard technique to factor out the highest power of \( x \) from the expression inside the fractional power.
Step 2: Key Formula or Approach:
Factor out \( x^3 \) from \( (x - x^3)^{1/3} \).
Then use substitution to integrate.
Step 3: Detailed Explanation:
Let \( I = \int_{1/3}^{1} \frac{(x - x^3)^{1/3}}{x^4} dx \).
Factor \( x^3 \) inside the parentheses:
\[ I = \int_{1/3}^{1} \frac{(x^3(\frac{1}{x^2} - 1))^{1/3}}{x^4} dx \]
\[ I = \int_{1/3}^{1} \frac{(x^3)^{1/3} \cdot (\frac{1}{x^2} - 1)^{1/3}}{x^4} dx \]
\[ I = \int_{1/3}^{1} \frac{x \cdot (\frac{1}{x^2} - 1)^{1/3}}{x^4} dx \]
Simplify the fraction:
\[ I = \int_{1/3}^{1} \frac{(\frac{1}{x^2} - 1)^{1/3}}{x^3} dx \]
Now, use substitution. Let \( t = \frac{1}{x^2} - 1 = x^{-2} - 1 \).
Differentiate with respect to \( x \):
\[ dt = -2x^{-3} dx \implies dt = -\frac{2}{x^3} dx \implies \frac{dx}{x^3} = -\frac{dt}{2} \]
Change the limits of integration:
When \( x = \frac{1}{3} \), \( t = \frac{1}{(1/3)^2} - 1 = \frac{1}{1/9} - 1 = 9 - 1 = 8 \).
When \( x = 1 \), \( t = \frac{1}{1^2} - 1 = 1 - 1 = 0 \).
Substitute everything into the integral:
\[ I = \int_{8}^{0} t^{1/3} \left(-\frac{dt}{2}\right) \]
Swap the limits to remove the negative sign:
\[ I = \frac{1}{2} \int_{0}^{8} t^{1/3} dt \]
Integrate:
\[ I = \frac{1}{2} \left[ \frac{t^{(1/3) + 1}}{(1/3) + 1} \right]_{0}^{8} = \frac{1}{2} \left[ \frac{t^{4/3}}{4/3} \right]_{0}^{8} \]
\[ I = \frac{1}{2} \cdot \frac{3}{4} \left[ t^{4/3} \right]_{0}^{8} = \frac{3}{8} [8^{4/3} - 0] \]
Evaluate \( 8^{4/3} \):
\[ 8^{4/3} = (8^{1/3})^4 = 2^4 = 16 \]
\[ I = \frac{3}{8} \cdot 16 = 3 \cdot 2 = 6 \]
Step 4: Final Answer:
The value of the integral is 6.