Answer
\[ = 2\]
Work Step by Step
\[\begin{gathered}
\int_0^4 {\frac{p}{{\sqrt {9 + {p^2}} }}\,\,\,\,dp} \hfill \\
\hfill \\
set\,\,u = 9 + {p^2}\,\,\,\,then\,\,\,du = 2pdp \hfill \\
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{\text{Changing limits of integration}} \hfill \\
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P = 0\,\,\,\,implies\,\,\,\,u = 9 \hfill \\
P = 4\,\,\,\,implies\,\,\,u = 25 \hfill \\
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Therefore, \hfill \\
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= \frac{1}{2}\int_9^{25} {\frac{{du}}{{{u^{\frac{1}{2}}}}}} \hfill \\
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integrate \hfill \\
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= \,\,\left[ {\frac{1}{2}\,\left( {2{u^{\frac{1}{2}}}} \right)} \right]_9^{25} \hfill \\
\hfill \\
= \,\,\left[ {\sqrt u } \right]_9^{25} \hfill \\
\hfill \\
Fundamental\,\,theorem \hfill \\
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= \sqrt {25} - \sqrt 9 \hfill \\
\hfill \\
Simplify \hfill \\
\hfill \\
= 2 \hfill \\
\end{gathered} \]