Question

Evaluate the integral.

\(\int\limits_{\rm{1}}^{\rm{4}} {\left( {\frac{{{\rm{4 + 6u}}}}{{\sqrt {\rm{u}} }}} \right){\rm{du}}} \)

Short answer

The value of the integral is \({\rm{36}}\)

Step-by-step solution

Rewrite the integral.

\(\begin{aligned}{c}\int\limits_{\rm{1}}^{\rm{4}} {\frac{{{\rm{4 + 6u}}}}{{\sqrt {\rm{u}} }}{\rm{du &=& }}\int\limits_{\rm{1}}^{\rm{4}} {\left( {\frac{{{\rm{4u}}}}{{\sqrt {\rm{u}} }}{\rm{ + }}\frac{{{\rm{6u}}}}{{\sqrt {\rm{u}} }}} \right){\rm{du &=& }}\int\limits_{\rm{1}}^{\rm{4}} {\left( {{\rm{4}}{{\rm{u}}^{{\raise0.7ex\hbox{${{\rm{ - 1}}}$} \!\mathord{\left/

{\vphantom {{{\rm{ - 1}}} {\rm{2}}}}\right.\kern-\nulldelimiterspace}

\!\lower0.7ex\hbox{${\rm{2}}$}}}}{\rm{ + 6}}{{\rm{u}}^{{\raise0.7ex\hbox{${\rm{1}}$} \!\mathord{\left/

{\vphantom {{\rm{1}} {\rm{2}}}}\right.\kern-\nulldelimiterspace}

\!\lower0.7ex\hbox{${\rm{2}}$}}}}} \right){\rm{du}}} } } \left( {{\rm{8}}{{\rm{u}}^{{\raise0.7ex\hbox{${\rm{1}}$} \!\mathord{\left/

{\vphantom {{\rm{1}} {\rm{2}}}}\right.\kern-\nulldelimiterspace}

\!\lower0.7ex\hbox{${\rm{2}}$}}}}{\rm{ + 4}}{{\rm{u}}^{{\raise0.7ex\hbox{${\rm{3}}$} \!\mathord{\left/

{\vphantom {{\rm{3}} {\rm{2}}}}\right.\kern-\nulldelimiterspace}

\!\lower0.7ex\hbox{${\rm{2}}$}}}}} \right)_{\rm{1}}^{\rm{4}}{\rm{ &=& }}\left( {{\rm{8(4}}{{\rm{)}}^{{\raise0.7ex\hbox{${\rm{1}}$} \!\mathord{\left/

{\vphantom {{\rm{1}} {\rm{2}}}}\right.\kern-\nulldelimiterspace}

\!\lower0.7ex\hbox{${\rm{2}}$}}}}{\rm{ + 4(4}}{{\rm{)}}^{{\raise0.7ex\hbox{${\rm{3}}$} \!\mathord{\left/

{\vphantom {{\rm{3}} {\rm{2}}}}\right.\kern-\nulldelimiterspace}

\!\lower0.7ex\hbox{${\rm{2}}$}}}}} \right){\rm{ - }}\left( {{\rm{8(1}}{{\rm{)}}^{{\raise0.7ex\hbox{${\rm{1}}$} \!\mathord{\left/

{\vphantom {{\rm{1}} {\rm{2}}}}\right.\kern-\nulldelimiterspace}

\!\lower0.7ex\hbox{${\rm{2}}$}}}}{\rm{ + 4(1}}{{\rm{)}}^{{\raise0.7ex\hbox{${\rm{3}}$} \!\mathord{\left/

{\vphantom {{\rm{3}} {\rm{2}}}}\right.\kern-\nulldelimiterspace}

\!\lower0.7ex\hbox{${\rm{2}}$}}}}} \right)\\{\rm{ &=& }}\left( {{\rm{16 + 32}}} \right){\rm{ - }}\left( {{\rm{8 + 4}}} \right)\\{\rm{ &=& 36}}\end{aligned}\)

Formula.

In step 1 we have used an equation

Such that,

\(\left( {{\rm{8(4}}{{\rm{)}}^{{\raise0.7ex\hbox{${\rm{1}}$} \!\mathord{\left/

{\vphantom {{\rm{1}} {\rm{2}}}}\right.\kern-\nulldelimiterspace}

\!\lower0.7ex\hbox{${\rm{2}}$}}}}{\rm{ + 4(4}}{{\rm{)}}^{{\raise0.7ex\hbox{${\rm{3}}$} \!\mathord{\left/

{\vphantom {{\rm{3}} {\rm{2}}}}\right.\kern-\nulldelimiterspace}

\!\lower0.7ex\hbox{${\rm{2}}$}}}}} \right)_{\rm{1}}^{\rm{4}}\) \(\left( {{\rm{Since}}\int {{{\rm{x}}^{\rm{n}}}} {\rm{dx = }}\frac{{{{\rm{x}}^{{\rm{n + 1}}}}}}{{{\rm{n + 1}}}}} \right)\)

Thus, the value of the integral \(\int\limits_{\rm{1}}^{\rm{4}} {\left( {\frac{{{\rm{4 + 6u}}}}{{\sqrt {\rm{u}} }}} \right){\rm{du}}} \) is \({\rm{36}}\)