MHB Calculate Vector Integral $\vec{V_2}$

[karush]

\begin{align*}\displaystyle
\vec{V_2}
&=\int_0^3 \left[\left(
\frac{4}{\sqrt{1+t}}\right){I}-\left(7t^2 \right){j}
+\left(\frac{14t}{\left(1+t^2 \right)^2}\right){k}\right] dt \\
&=\left[\int_0^3
\frac{4}{\sqrt{1+t}}
dt \right] {I}
-\left[\int_0^3 7t^2 dt\right] {j}
+\left[
\int_0^3\frac{14t}{\left(1+t^2 \right)^2}\, dt
\right]{k}
\end{align*}

Just seeing if going in the right direction

 

[Prove It]

Re: 13.2.2 vector integral

\begin{align*}\displaystyle
\vec{V_2}
&=\int_0^3 \left[\left(
\frac{4}{\sqrt{1+t}}\right){I}-\left(7t^2 \right){j}
+\left(\frac{14t}{\left(1+t^2 \right)^2}\right){k}\right] dt \\
&=\left[\int_0^3
\frac{4}{\sqrt{1+t}}
dt \right] {I}
-\left[\int_0^3 7t^2 dt\right] {j}
+\left[
\int_0^3\frac{14t}{\left(1+t^2 \right)^2}\, dt
\right]{k}
\end{align*}

Just seeing if going in the right direction

What is the actual question you are trying to solve?

 

[karush]

Re: 13.2.2 vector integral

\begin{align*}\displaystyle
\vec{V_2}
&=\int_0^3 \left[\left(
\frac{4}{\sqrt{1+t}}\right){I}-\left(7t^2 \right){j}
+\left(\frac{14t}{\left(1+t^2 \right)^2}\right){k}\right] dt \\
&=\left[\int_0^3
\frac{4}{\sqrt{1+t}}dt \right] {I}
-\left[\int_0^3 7t^2 dt\right] {j}
+\left[\int_0^3\frac{14t}{\left(1+t^2 \right)^2}\, dt \right]{k}\\
&=8\sqrt{t+1}\Biggr|_0^3i-14t\Biggr|_0^3j-\dfrac{7}{t^2+1} \Biggr|_0^3k\\
&=16i-52j-\frac{7}{10}k
\end{align*}

ok I made ? somewhere because the answer is supposed to be
$$8i-63j+\frac{63}{10}k$$

 

[topsquark]

Re: 13.2.2 vector integral

\begin{align*}\displaystyle
\vec{V_2}
&=\int_0^3 \left[\left(
\frac{4}{\sqrt{1+t}}\right){I}-\left(7t^2 \right){j}
+\left(\frac{14t}{\left(1+t^2 \right)^2}\right){k}\right] dt \\
&=\left[\int_0^3
\frac{4}{\sqrt{1+t}}dt \right] {I}
-\left[\int_0^3 7t^2 dt\right] {j}
+\left[\int_0^3\frac{14t}{\left(1+t^2 \right)^2}\, dt \right]{k}\\
&=8\sqrt{t+1}\Biggr|_0^3i-14t\Biggr|_0^3j-\dfrac{7}{t^2+1} \Biggr|_0^3k\\
&=16i-52j-\frac{7}{10}k
\end{align*}

ok I made ? somewhere because the answer is supposed to be
$$8i-63j+\frac{63}{10}k$$

First integral: [math]8 ~ \sqrt{ t + 1}[/math] over the limits is [math]8\sqrt{4} - 8 \sqrt{1}[/math]. You didn't subtract the last term.

Second integral: You didn't integrate, you took a derivative!

Third integral: You did the same thing as the first. Remember to subtract!

-Dan

 

[HallsofIvy]

Re: 13.2.2 vector integral

Don't you just hate that the editor automatically changes you "i" to "I"?
(I had to go back and correct that first "i".)

 

[MarkFL]

Re: 13.2.2 vector integral

Don't you just hate that the editor automatically changes you "i" to "I"?
(I had to go back and correct that first "i".)

i don't get any auto-capping of a single i in my posts...

 

[karush]

Re: 13.2.2 vector integral

$\textsf{Evaluate the Integral:}$
\begin{align*}\displaystyle
\vec{V_2}
&=\int_0^3 \left[\left[
\frac{4}{\sqrt{1+t}}\right]\textbf{i}-\left[7t^2 \right]\textbf{j}
+\left[\frac{14t}{\left[1+t^2 \right]^2}\right]\textbf{k}\right] dt \\
&=\left[\int_0^3
\frac{4}{\sqrt{1+t}}dt \right] \textbf{i}
-7\left[\int_0^3 t^2 dt\right] \textbf{j}
+14\left[\int_0^3\frac{t}{\left[1+t^2 \right]^2}\, dt \right]\textbf{k}\\
&=8[\sqrt{t+1}]
\Biggr|_0^3\textbf{i}
-7\left[\frac{t^3}{3}\right]
\Biggr|_0^3\textbf{j}
-14\left[\frac{1}{2\left[t^2+1\right]}\right]
\Biggr|_0^3\textbf{k}\\
&=[16-8]\textbf{i}-[7][9]\textbf{j}+7\left[\frac{9}{10}\right]\textbf{k}\\
&=8\textbf{i}-63\textbf{j}+\frac{63}{10}\textbf{k}
\end{align*}

OK hosed off the errors
hopefully

 

[karush]

Re: 13.2.2 vector integral

i don't get any auto-capping of a single i in my posts...

why isn't "\vec{}" on the select menu.

or was the vector crowd considered outcasts?

 

[MarkFL]

Re: 13.2.2 vector integral

why isn't "\vec{}" on the select menu.

or was the vector crowd considered outcasts?

Under the "Geometry" section you can find the vector choices:

View attachment 7397

 

[karush]

Re: 13.2.2 vector integral

$\displaystyle \int\sqrt{4x^2+1}\ dx$
$u=2x\therefore\frac{1}{2}du=dx $
$a=1$
$\displaystyle \frac{1}{2} \int\sqrt{u^2+a^2}\ du$
$\displaystyle\frac{1}{2}\left[\frac{u}{2}\sqrt{u^2+a^2} +
\frac{a^2}{2}\ln\left| u+\sqrt{u^2+a^2}\right| \right]$
then back plug $u$ and $a$

sorry thot this was a new post
but is this correct?