Question

plzz give me ans...ques no 9

Answer 1

$\int_{ \frac{1}{3} }^{1} \frac{ {(x - {x}^{3} )}^{ \frac{1}{3} } }{ {x}^{4} } dx \\ = \int_{ \frac{1}{3} }^{1} \frac{ {( {x}^{3} ( \frac{1}{ {x}^{2} } - 1) )}^{ \frac{1}{3} } }{ {x}^{4} } dx \\ = \int_{ \frac{1}{3} }^{1} \frac{ { x ( \frac{1}{ {x}^{2} } - 1 )}^{ \frac{1}{3} } }{ {x}^{4} } dx \\ = \int_{ \frac{1}{3} }^{1} \frac{ { ( \frac{1}{ {x}^{2} } - 1 )}^{ \frac{1}{3} } }{ {x}^{3} } dx \\ \\ let \: \: \frac{1}{ {x}^{2} } = y \\ then \: \: \frac{1}{ {x}^{3} } dx = \frac{ dy}{ - 2} \\ so \: \: integral \: \: becomes \\ \int_{ 9}^{1} \frac{ { ( y - 1 )}^{ \frac{1}{3} } }{ - 2} dy \\ = \frac{1}{2} \int_{ 1}^{9} {( y - 1 )}^{ \frac{1}{3} } dy \\ = \frac{1}{2} \times ( \frac{ {(y - 1)}^{ \frac{1}{3} + 1 } }{ \frac{1}{3} + 1} ) \: \: from \: \: 1 \: \: to \: \: 9 \\ = \frac{1}{2} \times ( \frac{ {(y - 1)}^{ \frac{4}{3} } }{ \frac{4}{3} } ) \: \: from \: \: 1 \: \: to \: \: 9 \\ = 6 - 0 = 6$

Option: (A)