Question

Evaluate the integral\(\int\limits_0^1 {\frac{{{r^3}}}{{\sqrt {4 + {r^2}} }}} dr\)

Short answer

We will use themethod of substitution to solve the given integral

Step-by-step solution

Given Data

Given =\({\rm I} = \int\limits_0^1 {\frac{{{r^3}}}{{\sqrt {4 + {r^2}} }}} dr\)

Substitute: \(\begin{aligned}{l}x = 4 + {r^2}\\{r^2} = x - 4\\dx = 2rdr\end{aligned}\)

\({\rm I} = \int\limits_0^1 {\frac{{{r^3}}}{{\sqrt {4 + {r^2}} }}} dr\)

solve the integration

\(\begin{aligned}{l}{\rm I} = \frac{1}{2}\int {\left( {\frac{x}{{\sqrt x }} - \frac{4}{{\sqrt x }}} \right)dx} \\ = \frac{1}{2}\int {\left( {\sqrt x - \frac{4}{{\sqrt x }}} \right)dx} \\ = \frac{1}{2}\left( {\frac{{{x^{{\raise0.7ex\hbox{$3$} \!\mathord{\left/

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

\!\lower0.7ex\hbox{$2$}}}}}}{{{\raise0.7ex\hbox{$3$} \!\mathord{\left/

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

\!\lower0.7ex\hbox{$2$}}}} - \frac{{4\sqrt x }}{{{\raise0.7ex\hbox{$1$} \!\mathord{\left/

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

\!\lower0.7ex\hbox{$2$}}}}} \right)\\ = \frac{{{x^{{\raise0.7ex\hbox{$3$} \!\mathord{\left/

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

\!\lower0.7ex\hbox{$2$}}}}}}{3} - 4\sqrt x \end{aligned}\)

Substitute back\(x = 4 + {r^2}\)

\(\)\({\rm I} = \left( {\frac{{{{\left( {{r^2} + 4} \right)}^{{\raise0.7ex\hbox{$3$} \!\mathord{\left/

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

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

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

\!\lower0.7ex\hbox{$2$}}}}} \right)} \right)_0^1\)

Substitute Limits

\(\begin{aligned}{l}{\rm I} = \left( {\frac{{{{\left( {{1^2} + 4} \right)}^{{\raise0.7ex\hbox{$3$} \!\mathord{\left/

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

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

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

\!\lower0.7ex\hbox{$2$}}}}} \right)} \right) - \left( {\frac{{{{\left( {{0^2} + 4} \right)}^{{\raise0.7ex\hbox{$3$} \!\mathord{\left/

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

\!\lower0.7ex\hbox{$2$}}}}}}{3} - 4\left( {{{\left( {{0^2} + 4} \right)}^{{\raise0.7ex\hbox{$1$} \!\mathord{\left/

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

\!\lower0.7ex\hbox{$2$}}}}} \right)} \right)\\ = \left( {\frac{{{5^{{\raise0.7ex\hbox{$3$} \!\mathord{\left/

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

\!\lower0.7ex\hbox{$2$}}}}}}{3} - 4\sqrt 5 } \right) - \left( {\frac{{{4^{{\raise0.7ex\hbox{$3$} \!\mathord{\left/

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

\!\lower0.7ex\hbox{$2$}}}}}}{3} - 4\sqrt 4 } \right)\\ = \frac{{ - 7}}{3}\sqrt 5 - \left( {\frac{8}{3} - 8} \right)\\ = \frac{{ - 7\sqrt 5 }}{3} - \frac{{8 - 24}}{3}\\ = \frac{{ - 7\sqrt 5 }}{3} + \frac{{16}}{3}\end{aligned}\)

Hence, \(\int\limits_0^1 {\frac{{{r^3}}}{{\sqrt {4 + {r^2}} }}} dr = \frac{1}{3}\left( {16 - 7\sqrt 5 } \right)\)