Evaluate the following integrals: _ 1 3 ^ 1 (x-x^3 )^ 1 3 x^4 dx - Vidyadip

Question

Evaluate the following integrals:
$\int_\limits{\frac{1}{3}}^{1}\frac{\big(\text{x}-\text{x}^3\big)^{\frac{1}{3}}}{\text{x}^4}\text{ dx}$

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Answer

Let $\text{I}=\int_\limits{\frac{1}{3}}^{1}\frac{\big(\text{x}-\text{x}^3\big)^{\frac{1}{3}}}{\text{x}^4}\text{ dx}$
$=\int_\limits{\frac{1}{3}}^{1}\frac{\Bigg[\text{x}^3\Big(\frac{\text{x}}{\text{x}^3}-1\Big)\Bigg]^{\frac{1}{3}}}{\text{x}^4}\text{ dx}$
$=\int_\limits{\frac{1}{3}}^{1}\frac{\text{x}\big(\frac{1}{\text{x}^2}-1\big)^{\frac{1}{3}}}{\text{x}^4}\text{ dx}$
$=\int_\limits{\frac{1}{3}}^{1}\frac{\text{x}\big(\frac{1}{\text{x}^2}-1\big)^{\frac{1}{3}}}{\text{x}^3}\text{ dx}$
Put $\Big(\frac{1}{\text{x}^2}-1\Big)=\text{Z}$
$\therefore\ -\frac{2}{\text{x}^3}\text{ dx}=\text{dz}$
$\Rightarrow\frac{\text{dx}}{\text{x}^3}=-\frac{\text{dz}}{2}$
When $\text{x}\rightarrow\frac{1}{3},\text{z}\rightarrow8$
When $\text{x}\rightarrow1,\text{z}\rightarrow0$
$\therefore\ \text{I}=-\frac{1}{2}\int^\limits0_8\text{z}^{\frac{1}{3}}\text{ dz}$
$=-\frac{1}{2}\times\Bigg[\frac{\text{z}^{\frac{4}{3}}}{\frac{4}{3}}\Bigg]^0_8$
$=-\frac{3}{8}\Big[0-(8)^{\frac{4}{3}}\Big]$
$=-\frac{3}{8}\times(-16)$
$=6$

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