Correct Answer - Option 3 :
\(\rm 32\over 15\)
Concept:
Definite Integral properties:
\(\mathop \smallint \limits_{\rm{a}}^{\rm{b}} {\rm{f}}\left( {\rm{x}} \right){\rm{\;dx}} = \mathop \smallint \limits_{\rm{a}}^{\rm{b}} {\rm{f}}\left( {{\rm{a}} + {\rm{b}} - {\rm{x}}} \right){\rm{\;dx}}\)
Calculation:
Given: f(x) = \(\rm \mathop \smallint \nolimits_0^2 {\rm{x}}{(2 - {\rm{x}})^4}{\rm{dx}}\)
Now using property, \(\mathop \smallint \limits_{\rm{a}}^{\rm{b}} {\rm{f}}\left( {\rm{x}} \right){\rm{\;dx}} = \mathop \smallint \limits_{\rm{a}}^{\rm{b}} {\rm{f}}\left( {{\rm{a}} + {\rm{b}} - {\rm{x}}} \right){\rm{\;dx}}\)
\(\rm \rm \mathop \smallint \nolimits_0^2 {\rm{x}}{(2 - {\rm{x}})^4}{\rm{dx}} = \mathop \smallint \limits_0^2 \left( {2 - {\rm{x}}} \right){\left\{ {2 - \left( {2 - {\rm{x}}} \right)} \right\}^4}{\rm{dx}}\)
\(\rm = \mathop \smallint \limits_0^2 \left( {2 - {\rm{x}}} \right){{\rm{x}}^4}{\rm{dx}}\)
\(= \mathop \smallint \limits_0^2 \left( {{{\rm 2{x}}^4} - {{\rm{x}}^{5}}} \right){\rm{dx}}\)
\(= \left[ {\frac{{{{\rm{2x}}^{5}}}}{{5}} - \frac{{{{\rm{x}}^{6}}}}{{6}}} \right]_0^2\)
= \(\rm {64\over 5} - {64\over 6}\)
= \(\rm 64\over 30\)
= \(\rm 32\over 15\)
