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Section 12.7 : Calculus with Vector Functions

8. Evaluate \(\displaystyle \int_{{ - 1}}^{2}{{\vec r\left( t \right)\,dt}}\) where \(\vec r\left( t \right) = \left\langle {6,6{t^2} - 4t,t{{\bf{e}}^{2t}}} \right\rangle \)

Show Solution

There really isn’t a lot to do here with this problem. All we need to do is integrate of all the components of the vector.

\[\begin{align*}\int{{\vec r\left( t \right)\,dt}} & = \left\langle {\int{{6\,dt}},\int{{6{t^2} - 4t\,dt}},\int{{t{{\bf{e}}^{2t}}\,dt}}} \right\rangle \\ & = \left\langle {\int{{6\,dt}},\int{{6{t^2} - 4t\,dt}},\frac{1}{2}t{{\bf{e}}^{2t}} - \frac{1}{2}\int{{{{\bf{e}}^{2t}}\,dt}}} \right\rangle = \left\langle {6t,2{t^3} - 2{t^2},\frac{1}{2}t{{\bf{e}}^{2t}} - \frac{1}{4}{{\bf{e}}^{2t}}} \right\rangle \end{align*}\]

Don’t forget the basic integration rules such integration by parts (third term). Also recall that one way to do definite integral is to first do the indefinite integral and then do the evaluation.

The answer for this problem is then,

\[\begin{align*}\int_{{ - 1}}^{2}{{\vec r\left( t \right)\,dt}} & = \left. {\left\langle {6t,2{t^3} - 2{t^2},\frac{1}{2}t{{\bf{e}}^{2t}} - \frac{1}{4}{{\bf{e}}^{2t}}} \right\rangle } \right|_{ - 1}^2\\ & = \left\langle {12,8,\frac{3}{4}{{\bf{e}}^4}} \right\rangle - \left\langle { - 6, - 4, - \frac{3}{4}{{\bf{e}}^{ - 2}}} \right\rangle = \require{bbox} \bbox[2pt,border:1px solid black]{{\left\langle {18,12,\frac{3}{4}\left( {{{\bf{e}}^4} + {{\bf{e}}^{ - 2}}} \right)} \right\rangle }}\end{align*}\]

We didn’t put a lot of the integration details into the solution on the assumption that you do know your integration skills well enough to follow what is going on. If you are somewhat rusty with your integration skills then you’ll need to go back to the integration material from both Calculus I and Calculus II to refresh your integration skills.