Question

The value of \( \int_{1/3}^{1} \frac{(x - x^3)^{\frac{1{3}}}{x^4} dx \) is

• A. 0
• B. 2
• C. 4
• D. 6

Hint:

When you see terms like \( (x-x^n)^{1/n} \), try factoring out \( x^n \).

Answer 1

The Correct Option is D

Step 1: Concept Use substitution to simplify the integral.

Step 2: Meaning Factor \( x^3 \) out of the parenthesis: \( (x^3(1/x^2 - 1))^{1/3} = x(1/x^2 - 1)^{1/3} \). Integral becomes \( \int \frac{(1/x^2 - 1)^{1/3}}{x^3} dx \).

Step 3: Analysis Let \( t = 1/x^2 - 1 \), then \( dt = -2/x^3 dx \). Limits: \( x = 1/3 \implies t = 8 \); \( x = 1 \implies t = 0 \). Integral = \( -1/2 \int_8^0 t^{1/3} dt = 1/2 \int_0^8 t^{1/3} dt \). Value = \( 1/2 [ \frac{3}{4}t^{4/3} ]_0^8 = \frac{3}{8}(8)^{4/3} = \frac{3}{8}(16) = 6 \).

Step 4: Conclusion The value of the definite integral is 6. Final Answer: (D)