Question

Evaluate the integral.

\(\) \(\int\limits_{\rm{0}}^{\rm{4}} {{\rm{(3}}\sqrt {\rm{t}} {\rm{ - 2}}{{\rm{e}}^{\rm{t}}}{\rm{)dt}}} \)

Short answer

The value of the integral is \(\) \({\rm{18 - 2}}{{\rm{e}}^{\rm{4}}}\)

Step-by-step solution

Square root as a fractional power, general antiderivative.

To rewrite the square root as a fractional power,

\(\int\limits_{\rm{0}}^{\rm{4}} {{\rm{(3}}\sqrt {\rm{t}} {\rm{ - 2}}{{\rm{e}}^{\rm{t}}}{\rm{)dt = }}\int\limits_{\rm{0}}^{\rm{4}} {{\rm{3}}{{\rm{t}}^{{\raise0.7ex\hbox{${\rm{1}}$} \!\mathord{\left/

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

\!\lower0.7ex\hbox{${\rm{2}}$}}}}} {\rm{ - 2}}{{\rm{e}}^{\rm{t}}}{\rm{)dt}}} \)

To find the general antideriavative, using the indefinite integrals

and \(\int {{{\rm{e}}^{\rm{t}}}{\rm{dt = }}{{\rm{e}}^{\rm{t}}}{\rm{ + C}}} \)

Ignore (just for the moment) the boundaries of integration.

Leave out arbitrary constant.

\(\begin{array}{c}\int {{\rm{(3}}{{\rm{t}}^{{\raise0.7ex\hbox{${\rm{1}}$} \!\mathord{\left/

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

\!\lower0.7ex\hbox{${\rm{2}}$}}}}{\rm{ - 2}}{{\rm{e}}^{\rm{t}}}{\rm{)dt = }}\int {{\rm{3}}{{\rm{t}}^{{\raise0.7ex\hbox{${\rm{1}}$} \!\mathord{\left/

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

\!\lower0.7ex\hbox{${\rm{2}}$}}}}} {\rm{dt - }}\int {{\rm{2}}{{\rm{e}}^{\rm{t}}}} {\rm{dt = 3}}\frac{{{{\rm{t}}^{{\raise0.7ex\hbox{${\rm{t}}$} \!\mathord{\left/

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

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

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

\!\lower0.7ex\hbox{${\rm{2}}$}}{\rm{ + 1}}}}{\rm{ - 2}}{{\rm{e}}^{\rm{t}}}{\rm{ + C}}} {\rm{ = 3 \times }}\frac{{{{\rm{t}}^{{\raise0.7ex\hbox{${\rm{3}}$} \!\mathord{\left/

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

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

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

\!\lower0.7ex\hbox{${\rm{2}}$}}}}{\rm{ - 2}}{{\rm{e}}^{\rm{t}}}{\rm{ + C}}\\{\rm{ = 3 \times }}\frac{{\rm{2}}}{{\rm{3}}}{{\rm{t}}^{{\raise0.7ex\hbox{${\rm{3}}$} \!\mathord{\left/

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

\!\lower0.7ex\hbox{${\rm{2}}$}}}}{\rm{ - 2}}{{\rm{e}}^{\rm{t}}}{\rm{ + C}}\\{\rm{ = 2}}{{\rm{t}}^{{\raise0.7ex\hbox{${\rm{3}}$} \!\mathord{\left/

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

\!\lower0.7ex\hbox{${\rm{2}}$}}}}{\rm{ - 2}}{{\rm{e}}^{\rm{t}}}{\rm{ + C}}\\{\rm{ = F(t)}}\end{array}\)

We will leave out the arbitrary constant C, since it cancels out when calculating a definite integral.

Assigning final value.

\(\begin{array}{c}\int\limits_{\rm{0}}^{\rm{4}} {{\rm{(3}}{{\rm{t}}^{{\raise0.7ex\hbox{${\rm{1}}$} \!\mathord{\left/

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

\!\lower0.7ex\hbox{${\rm{2}}$}}{\rm{ - }}}}} {\rm{2}}{{\rm{e}}^{\rm{t}}}{\rm{)dt = }}\left( {{\rm{2}}{{\rm{t}}^{{\raise0.7ex\hbox{${\rm{3}}$} \!\mathord{\left/

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

\!\lower0.7ex\hbox{${\rm{2}}$}}}}{\rm{ - 2}}{{\rm{e}}^{\rm{t}}}} \right)_{\rm{0}}^{\rm{4}}{\rm{ = }}\left( {{\rm{2(4}}{{\rm{)}}^{{{\rm{3}} \mathord{\left/

{\vphantom {{\rm{3}} {}}} \right.

\kern-\nulldelimiterspace} {}}{\rm{2}}}}{\rm{ - 2}}{{\rm{e}}^{\rm{4}}}} \right){\rm{ - }}\left( {{\rm{2(0}}{{\rm{)}}^{{{\rm{3}} \mathord{\left/

{\vphantom {{\rm{3}} {}}} \right.

\kern-\nulldelimiterspace} {}}{\rm{2}}}}{\rm{ - 2}}{{\rm{e}}^{\rm{0}}}} \right)\\{\rm{ = }}\left( {{\rm{16 - 2}}{{\rm{e}}^{\rm{4}}}} \right){\rm{ - }}\left( {{\rm{ - 2}}} \right)\\{\rm{ = 18 - 2}}{{\rm{e}}^{\rm{4}}}\end{array}\)