Question

\(\int\limits_0^2 {{x^3}\sqrt {{x^2} + 4} } dx\) Evaluate the integral

Short answer

After evaluation \(\int\limits_0^2 {{x^3}\sqrt {{x^2} + 4} } dx\),will be equal to \(\frac{{65}}{{14}}\left( {\sqrt 2 + 1} \right)\)

Step-by-step solution

Step-1: - (Substitute \({x^2} + 4 = t\) and solving)

Given \(\int\limits_0^2 {{x^3}\sqrt {{x^2} + 4} } dx\)

Let

Differentiating w.r.t x

\(\begin{aligned}{l}\frac{d}{{dx}}\left( {{x^2} + 4} \right) &= \frac{d}{{dx}}\left( t \right)\\2x &= \frac{{dt}}{{dx}}\\xdx &= \frac{{dt}}{2}\end{aligned}\)

So

Step-2: - (Solving the problem)

\(\begin{aligned}{l}I &= \frac{1}{2}\int\limits_2^8 {\left( {t - 4} \right)\sqrt t dt} \\I &= \frac{1}{2}\int\limits_4^8 {t\sqrt t dt - } \frac{1}{2}\int\limits_4^8 {4\sqrt t dt} \\I &= \frac{1}{2}\int\limits_4^8 {{t^{{\raise0.7ex\hbox{$3$} \!\mathord{\left/

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

\!\lower0.7ex\hbox{$2$}}}}} dt - 2\int\limits_4^8 {\sqrt t } dt\\I &= \frac{1}{2}\frac{2}{5}\left| {{t^{{\raise0.7ex\hbox{$5$} \!\mathord{\left/

{\vphantom {5 2}}\right.\kern-\nulldelimiterspace}

\!\lower0.7ex\hbox{$2$}}}}} \right|_4^8 - 2\frac{2}{3}\left| {{t^{{\raise0.7ex\hbox{$8$} \!\mathord{\left/

{\vphantom {8 2}}\right.\kern-\nulldelimiterspace}

\!\lower0.7ex\hbox{$2$}}}}} \right|_4^8\\I &= \frac{1}{5}\left| {{5^{{\raise0.7ex\hbox{$5$} \!\mathord{\left/

{\vphantom {5 2}}\right.\kern-\nulldelimiterspace}

\!\lower0.7ex\hbox{$2$}}}} - {4^{{\raise0.7ex\hbox{$5$} \!\mathord{\left/

{\vphantom {5 2}}\right.\kern-\nulldelimiterspace}

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

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

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

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

\!\lower0.7ex\hbox{$2$}}}}} \right|\end{aligned}\)

Step-3: (Simplifying like terms)

Hence after evaluation \(\int\limits_0^2 {{x^3}\sqrt {{x^2} + 4} } dx\) , will be equal to \(\frac{{64}}{{15}}\left( {\sqrt 2 + 1} \right)\)