Compute the indefinite integral of the following function. r(t)

jeseHainsij 2021-11-17 Answered
Compute the indefinite integral of the following function.
\(\displaystyle{r}{\left({t}\right)}={14}^{{{t}}}{i}+{\frac{{{1}}}{{{3}+{8}{t}}}}{j}+{\ln{{19}}}{t}{k}\)
Select the correct choice below and fill in the answer boxes to complete your choice.
A) \(\displaystyle\int{r}{\left({t}\right)}{\left.{d}{t}\right.}={()}{i}+{()}{j}+{()}{k}\)
B) \(\displaystyle\int{r}{\left({t}\right)}{\left.{d}{t}\right.}={()}{i}+{()}{j}+{()}{k}+{C}\)

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Expert Answer

May Dunn
Answered 2021-11-18 Author has 456 answers
Step 1
Solution:
Consider the vector-valued function:
\(\displaystyle{r}{\left({t}\right)}={14}^{{{t}}}{i}+{\frac{{{1}}}{{{3}+{8}{t}}}}{j}+{\ln{{19}}}{t}{k}\)
Integrating with respect to t we have
\(\displaystyle\int{r}{\left({t}\right)}={\left[\int{\left({14}\right)}^{{{t}}}{\left.{d}{t}\right.}\right]}{i}+{\left[\int{\left({\frac{{{1}}}{{{3}+{8}{t}}}}\right)}{\left.{d}{t}\right.}\right]}+{\left[\int{\left({\ln{{19}}}{t}\right)}{\left.{d}{t}\right.}\right]}{k}\)
\(\displaystyle{\sin{{c}}}{e}\int{\left({14}\right)}^{{{t}}}{\left.{d}{t}\right.}={\frac{{{14}^{{{t}}}}}{{{\ln{{14}}}}}}\)
\(\displaystyle\int{\left({\frac{{{1}}}{{{3}+{8}{t}}}}\right)}{\left.{d}{t}\right.}={\frac{{{1}}}{{{8}}}}\int{\frac{{{1}}}{{{u}}}}{d}{u}{\left({a}{s}\ {p}{u}{t}\ {u}={3}+{8}{t}\right)}\)
\(\displaystyle\int{\left({\frac{{{1}}}{{{3}+{8}{t}}}}\right)}{\left.{d}{t}\right.}={\frac{{{1}}}{{{8}}}}{\ln}{\left|{u}\right|}={\frac{{{1}}}{{{8}}}}{\left|{3}+{8}{t}\right|}\) and
\(\displaystyle\int{\left({\ln{{19}}}{t}\right)}{\left.{d}{t}\right.}={\left({t}{\left({\ln{{19}}}{t}\right)}-{t}\right)}\)
Putting these values in above we have
\(\displaystyle\int{r}{\left({t}\right)}={\left({\frac{{{14}^{{{t}}}}}{{{\ln{{14}}}}}}\right)}{i}+{\left({\frac{{{1}}}{{{8}}}}{\ln}{\left|{3}+{8}{t}\right|}\right)}{j}+{\left({t}{\left({\ln{{19}}}{t}\right)}-{t}\right)}{k}+{C}\)
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