Question

Choose the correct answer.
The value of integral
${\int }_{\frac{1}{3}}^1\frac{(x-x^3{)}^{\frac{1}{3}}}{x^4}dx$ is
(a)6
(b)0
(c)3
(d)4

Answer 1

Let $I={\int }_{\frac{1}{3}}^1\frac{(x-x^3{)}^{\frac{1}{3}}}{x^4}dx{\int }_{\frac{1}{3}}^1\frac{(x^3{)}^{\frac{1}{3}(\frac{x}{x^3}-\frac{x^3}{x^3}{)}^{\frac{1}{3}}}}{x^4}dx$
[Multiply and divide the numerator by $\left(x^3{)}^{\frac{1}{3}}\right]={\int }_{\frac{1}{3}}^1\frac{(\frac{1}{x^2}-1{)}^{\frac{1}{3}}}{x^3}dx$
Put $\frac{1}{x^2}=t\Rightarrow \frac{-2}{x}dx=dt\Rightarrow \frac{dx}{x^3}=\frac{dt}{(-2)}$
For limit when $x=1\Rightarrow \text{t =1 and when x}=\frac{1}{3}\Rightarrow t=3^2=9(\because t=\frac{1}{x^2})$

$\therefore I={\int }_9^1(t-1{)}^{\frac{1}{3}}\frac{dt}{(-2)}=-\frac{1}{2}{\left[\frac{(t-1{)}^{\frac{1}{3}+1}}{\frac{1}{3}+1}\right]}_9^1=-\frac{1}{2}\times \frac{3}{4}{\left[\right(t-1{)}^{\frac{4}{3}}]}_9^1=-\frac{3}{8}\left[\right(1-1{)}^{\frac{4}{3}}-(9-1{)}^{\frac{4}{3}}]=-\frac{3}{8}[0-(2^3{)}^{\frac{4}{3}}]=-\frac{3}{8}\times (-16)=6$
Hence, the option (a) is correct.