Answer
\[ = 2\ln 2 - 1\]
Work Step by Step
\[\begin{gathered}
\int_0^{\ln 2} {x{e^x}dx} \hfill \\
\hfill \\
set\,\,\,the\,\,substitution \hfill \\
\hfill \\
u = x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \to \,\,\,\,\,\,du = dx \hfill \\
dv = {e^x}dx\,\,\,\,\, \to \,\,\,\,\,\,\,v = {e^x} \hfill \\
\hfill \\
use\,\,uv - \int_{}^{} {vdu} \hfill \\
\hfill \\
{\text{replacing the values }}{\text{in the equation}} \hfill \\
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= x{e^x} - \int_{}^{} {{e^x}dx} \hfill \\
\hfill \\
integrate \hfill \\
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x{e^x} - {e^x} + C \hfill \\
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Therefore \hfill \\
\hfill \\
\int_0^{\ln 2} {x{e^x}dx} = \,\,\left[ {x{e^x} - {e^x}} \right]_0^{\ln 2} \hfill \\
\hfill \\
evaluate\,\,the\,\,limits \hfill \\
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= \,\left( {\ln 2{e^{\ln 2}} - {e^{\ln 2}}} \right) - \,\left( {0 - {e^0}} \right) \hfill \\
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= 2\ln 2 - 2 + 1 \hfill \\
\hfill \\
simplify \hfill \\
\hfill \\
= 2\ln 2 - 1 \hfill \\
\hfill \\
\end{gathered} \]
