${\int }_C\mathbf{F}·\mathbf{dr}$

$\={\int }_a^{b}f \mathrm{dx}\+g \mathrm{dy}$

$\={\int }_0^1\left[\left(\genfrac{}{}{0ex}{}{x y}{\phantom{x\=a}}|\genfrac{}{}{0ex}{}{\phantom{\mathrm{f(x)}}}{x\=x\left(t\right)\,y\=y\left(t\right)}\right)\cdot \frac{d}{\mathrm{dt}}x\left(t\right)\+\left(\genfrac{}{}{0ex}{}{x\+y}{\phantom{x\=a}}|\genfrac{}{}{0ex}{}{\phantom{\mathrm{f(x)}}}{x\=x\left(t\right)\,y\=y\left(t\right)}\right)\cdot \frac{d}{\mathrm{dt}}y\left(t\right)\right] \mathrm{dt}$

$\={\int }_0^1\left[\left(1\+2 t\right)\left(2\+3 t\right)\cdot 2\+\left(\left(1\+2 t\right)\+\left(2\+3 t\right)\right)\cdot 3\right] \mathrm{dt}$

$\={\int }_0^1\left(12t^2\+29t\+13\right) \mathrm{dt}$

$\=63\/2$

Initialize

Loading Student:-VectorCalculus

Define the vector field F

$⟨x y\,x\+y⟩$ = $\left[\begin{array}{c}xy\\ x+y\end{array}\right]$$\stackrel{\text{to Vector Field}}{\to }$$\left[\begin{array}{c}xy\\ x+y\end{array}\right]$$\stackrel{\text{assign to a name}}{\to }$$F$$$

Obtain the line integral via the Context Panel

$\mathbf{F}$

$\stackrel{\text{line integral}}{\to }$

${\∫}_0^1\left(2\left(1\+2t\right)\left(2\+3t\right)\+9\+15t\right)\ⅆt$

$\stackrel{\text{simplify}}{\=}$

${\∫}_0^1\left(12t^2\+29t\+13\right)\ⅆt$

$\=$

$\frac{63}{2}$

Define the path parametrically, as a position vector

$P\=\left[⟨1\,2⟩\,⟨3\,5⟩\right]$$\stackrel{\text{assign}}{\to }$

$\mathbf{R}\={\mathbf{P}}_1\+t \left({\mathbf{P}}_2-{\mathbf{P}}_1\right)$$\stackrel{\text{assign}}{\to }$

Obtain T, a unit tangent vector along the path

$\mathbf{R}$ = $\stackrel{\text{tangent vector}}{\to }$ $\stackrel{\text{Euclidean-normalize}}{\to }$ $\stackrel{\text{assign to a name}}{\to }$$T$$$

Obtain $\mathrm{\ρ}\=\frac{\mathrm{ds}}{\mathrm{dt}}\=∥\stackrel{\.}{\mathbf{R}}∥$ 

$∥\frac{\ⅆ}{\ⅆ t} \mathbf{R}∥$ = $\sqrt{13}$$\stackrel{\text{assign to a name}}{\to }$$\mathrm{\ρ}$$$

Obtain the integrand $\mathbf{F}·\mathbf{T} \mathrm{ds}$ 

$\mathbf{F}·\mathbf{T} \mathrm{\ρ}$

$\left(\frac{2}{13}\left(1\+2t\right)\left(2\+3t\right)\sqrt{13}\+\frac{3}{13}\left(3\+5t\right)\sqrt{13}\right)\sqrt{13}$

$\stackrel{\text{simplify}}{\=}$

$12t^2\+29t\+13$

$\stackrel{\text{integrate w.r.t. t}}{\to }$

${\∫}_0^1\left(12t^2\+29t\+13\right)\ⅆt$

$\=$

$\frac{63}{2}$

Initialize

$$\begin{gathered} \mathrm{with}\left(\mathrm{Student}:-\mathrm{VectorCalculus}\right)\: \\ \mathrm{BasisFormat}\left(\mathrm{false}\right)\: \end{gathered}$$

$\mathbf{F}≔\mathrm{VectorField}\left(⟨x y\,x\+y⟩\right)\:$

Apply the LineInt command

$\mathrm{LineInt}\left(\mathbf{F}\,\mathrm{Line}\left(⟨1\,2⟩\,⟨3\,5⟩\right)\,\mathrm{output}\=\mathrm{integral}\right)\=\mathrm{LineInt}\left(\mathbf{F}\,\mathrm{Line}\left(⟨1\,2⟩\,⟨3\,5⟩\right)\right)$

${\∫}_0^1\left(2\left(1\+2t\right)\left(2\+3t\right)\+9\+15t\right)\ⅆt\=\frac{63}{2}$

Use the LineInt command to generate a graph of F and the path of integration

$\mathrm{LineInt}\left(\mathbf{F}\,\mathrm{Line}\left(⟨1\,2⟩\,⟨3\,5⟩\right)\,\mathrm{output}\=\mathrm{plot}\right)$

Define the path parametrically, as the position vector R

$$\begin{gathered} L≔\left[⟨1\,2⟩\,⟨3\,5⟩\right]\: \\ \mathbf{R}≔{\mathbf{L}}_1\+t \left({\mathbf{L}}_2-{\mathbf{L}}_1\right) \end{gathered}$$

Obtain $\mathrm{\ρ}\=\frac{\mathrm{ds}}{\mathrm{dt}}\=∥\stackrel{\.}{\mathbf{R}}∥$ 

$\mathrm{\ρ}≔\mathrm{Norm}\left(\mathrm{diff}\left(\mathbf{R}\,t\right)\right)$

$\sqrt{13}$

Obtain T, a unit tangent vector along the path

$\mathbf{T}≔\mathrm{TangentVector}\left(\mathbf{R}\,t\,\mathrm{normalized}\right)$

Form and evaluate the requisite line integral

$\mathrm{Int}\left(\mathrm{DotProduct}\left(\mathbf{F}\,\mathbf{T}\right)\cdot \mathrm{\ρ}\,t\=0..1\right)\=\mathrm{int}\left(\mathrm{DotProduct}\left(\mathbf{F}\,\mathbf{T}\right)\cdot \mathrm{\rho }\,t\=0..1\right)$

${\∫}_0^1\left(\frac{2}{13}\left(1\+2t\right)\left(2\+3t\right)\sqrt{13}\+\frac{3}{13}\left(3\+5t\right)\sqrt{13}\right)\sqrt{13}\ⅆt\=\frac{63}{2}$