\[\mathbf{F}\]
and a curve \[C\]
the integral of the component of \[\mathbf{F}\]
normal to \[C\]
along \[C\]
from \[A\]
to \[B\]
is given by \[\int^B_A \mathbf{F_n} \cdot d \mathbf{r}\]
where \[\mathbf{F_n}\]
is the component of \[\mathbf{F}\]
normal to \[C\]
.In two the
\[xy\]
plane we can take \[\mathbf{n} = \mathbf{T} \times \mathbf{k}=(\frac{dx}{ds} \mathbf{i} + \frac{dy}{ds}\mathbf{j}) \times \mathbf{k}=\frac{dy}{ds}\mathbf{i} -\frac{dx}{ds}\mathbf{j} \]
.With
\[\mathbf{F}=F_1 \mathbf{i} + F_2 \mathbf{j}\]
we have\[\int^B_A \mathbf{F_n} d \mathbf{r} =\int^B_A \mathbf{F} \cdot \mathbf{n} ds= \int^B_A -F_2 dx + F_1 dy\]
Take
\[A=(0,0),B=(1,4)\]
, take \[\mathbf{F}=e^x \mathbf{i} + e^{3x} \mathbf{j}\]
and take the curve \[C\]
to be \[y=x^2 +3x \rightarrow dy=2xdx+3dx=(2x+3)dx\]
.Then
\[\begin{equation} \begin{aligned} \int^B_A \mathbf{F_n} \cdot d \mathbf{r} &= \int^1_0 -e^{3x} dx + int^1_0 e^x (2x+3) dx \\ &= [- \frac{1}{3}e^{3x}]^1_0 +[(2x+3)e^x - 2e^x]^1_0 \\ &= - \frac{1}{3}(e^3 -1)+ 3e-1 \\ &=3e-\frac{1}{3}e^3 - \frac{2}{3} \end{aligned} \end{equation}\]