Question

Evaluate: ${\int }_{1/3}^1\frac{(x-x^3{)}^{1/3}}{x^4}dx$.

• A. $3$
• B. $0$
• C. $6$
• D. $4$

Answer 1

Answer: (C)

$6$

${\int }_{\frac{1}{3}}^1\frac{(x–x^3{)}^{\frac{1}{3}}}{x^4}dx={\int }_{\frac{1}{3}}^1\frac{(x^3{)}^{\frac{1}{3}}{(\frac{1}{x^2}–1)}^{\frac{1}{3}}}{x^4}dx$

$={\int }_{\frac{1}{3}}^1\frac{{(\frac{1}{x^2}–1)}^{\frac{1}{3}}}{x^3}dx$ (let $\frac{1}{x^2}–1=t$)

$={\int }_8^0\frac{t^{\frac{1}{3}}}{-2}dt$ $\therefore \frac{-2}{x^3}dx=dt$

$=-\frac{1}{2}{\left(\frac{t^{4/3}}{4/3}\right)}_8^0=6$.