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Integral of Normal Component of a Vector Field Along a Curve

Give a vector field  
\[\mathbf{F}\]
  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
\[\begin{equation} \begin{aligned} \int^B_A \mathbf{F_n} \cdot d \mathbf{r} &= \int^1_0 -e^{3x} dx + int^1_0 e^x (2x+3) dx \\ &= [- \frac{1}{3}e^{3x}]^1_0 +[(2x+3)e^x - 2e^x]^1_0 \\ &= - \frac{1}{3}(e^3 -1)+ 3e-1 \\ &=3e-\frac{1}{3}e^3 - \frac{2}{3} \end{aligned} \end{equation}\]