Answer
\[ = \frac{7}{2}\]
Work Step by Step
\[\begin{gathered}
\int_0^1 {2x\,\left( {4 - {x^2}} \right)dx} \hfill \\
\hfill \\
let\,\,u = 4 - {x^2}\,\,\,\,{\text{which implies that}}\,\,du = - 2xdx \hfill \\
and\,\, - du = 2xdx \hfill \\
\hfill \\
apply\,\,the\,\,\,substitution \hfill \\
\hfill \\
= \int_{}^{} {u\,\left( { - du} \right)} \hfill \\
\hfill \\
integrate \hfill \\
\hfill \\
= \frac{{ - {u^2}}}{2} + C \hfill \\
\hfill \\
therefore \hfill \\
\hfill \\
= \frac{{ - \,{{\left( {4 - {x^2}} \right)}^2}}}{2} + C \hfill \\
\hfill \\
then \hfill \\
\hfill \\
= \,\,\left[ {\frac{{\,{{\left( {4x - {x^2}} \right)}^2}}}{2}} \right]_1^0 \hfill \\
\hfill \\
Fundamental\,\,theorem \hfill \\
\hfill \\
= \frac{{\,{{\left( {4 - {0^2}} \right)}^2}}}{2} - \frac{{\,{{\left( {4 - {1^2}} \right)}^2}}}{2} \hfill \\
\hfill \\
= \frac{7}{2} \hfill \\
\end{gathered} \]
