# Evaluate the line integral, where C is the given curve. integral C

Evaluate the line integral, where C is the given curve. integral C $x{e}^{yzds}$ C is the line segment from $\left(0,0,0\right)$ to $\left(1,2,3\right)$
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Jennifer Hill

Step 1
Let C the line segment from $\left(0,0,0\right)$ to $\left(1,2,3\right)$. The parametric equation for this line in vector form is
$r\left(t\right)={r}_{0}+td$
where we chose ${r}_{0}=⟨0,0,0⟩r$ and the direction vector d is
$d=⟨1,2,3⟩-⟨0,0,0⟩=⟨1,2,3⟩$
Hence we have
$r\left(t\right)=⟨t,2t,3t⟩$
and we have in scalar form the parametric equations
for C :
$x=t$, $y=2t$, $z=3t$, $0\le t\le 1$
To detemine the line integral:
$I={\int }_{C}x{e}^{yz}ds$
we first determine ds:
$ds=\sqrt{{\left(\frac{dx}{dt}\right)}^{2}+{\left(\frac{dx}{dt}\right)}^{2}+{\left(\frac{dz}{dt}\right)}^{2}}dt$
$=\sqrt{{\left(1\right)}^{2}+{\left(2\right)}^{2}={\left(3\right)}^{2}}dt$
$=\sqrt{1+4+9}dt$
$=\sqrt{14}dt$
Step 2
${\int }_{C}x{e}^{yz}ds={\int }_{0}^{1}t{e}^{6{t}^{2}}\sqrt{14}dt=\sqrt{14}{\int }_{0}^{1}{e}^{6{t}^{2}}tdt$
Now changing to the variable $u=6{t}^{2}$ , we have $du=12tdt=12tdt$ and:
$I=\sqrt{14}{\int }_{0}^{6}\frac{1}{12}{e}^{u}du=\frac{\sqrt{14}}{12}{\int }_{0}^{6}{e}^{u}du$
$=\frac{\sqrt{14}}{12}{e}^{u}\overline{)\begin{array}{c}{}^{6}\\ {}_{0}\end{array}}=\frac{\sqrt{14}}{12}\left({e}^{6}-{e}^{0}\right)$
$=\frac{\sqrt{14}}{12}\left({e}^{6}-1\right)$
Finally: