${\int }_C\mathbf{F}·\mathbf{dr}$

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

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

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

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

$\=63\/2$

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

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

$\=63\/2$

${\int }_0^1\left[\left(3\+t\right)\left(5-5 t\right)\cdot 1\+\left(\left(3\+t\right)\+\left(5-5 t\right)\right)\cdot \left(-5\right)\right] \mathrm{dt}$

$\=-5{\int }_0^1\left(t^2-2 t\+5\right) \mathrm{dt}$

$\=-65\/3$

${\int }_0^1\left[\left(4-2 t\right)\left(7 t\right)\cdot \left(-2\right)\+\left(\left(4-2 t\right)\+\left(7 t\right)\right)\cdot 7\right] \mathrm{dt}$

$\=7{\int }_0^1\left(4 t^2-3 t\+4\right) \mathrm{dt}$

$\=161\/6$

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

$\mathbf{F}$

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

${\∫}_0^1\left(2\left(1\+2t\right)\left(2\+3t\right)\+9\+15t\right)\ⅆt\+{\∫}_0^1\left(\left(3\+t\right)\left(5-5t\right)-40\+20t\right)\ⅆt\+{\∫}_0^1\left(-14\left(4-2t\right)t\+28\+35t\right)\ⅆt$

$\=$

$\frac{110}{3}$

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

$$\begin{gathered} \mathrm{LineInt}\left(\mathbf{F}\,\mathrm{LineSegments}\left(⟨1\,2⟩\,⟨3\,5⟩\,⟨4\,0⟩\,⟨2\,7⟩\right)\,\mathrm{output}\=\mathrm{integral}\right)\= \\ \mathrm{LineInt}\left(\mathbf{F}\,\mathrm{LineSegments}\left(⟨1\,2⟩\,⟨3\,5⟩\,⟨4\,0⟩\,⟨2\,7⟩\right)\right) \end{gathered}$$

${\∫}_0^1\left(2\left(1\+2t\right)\left(2\+3t\right)\+9\+15t\right)\ⅆt\+{\∫}_0^1\left(\left(3\+t\right)\left(5-5t\right)-40\+20t\right)\ⅆt\+{\∫}_0^1\left(-14\left(4-2t\right)t\+28\+35t\right)\ⅆt\=\frac{110}{3}$

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

$\mathrm{LineInt}\left(\mathbf{F}\,\mathrm{LineSegments}\left(⟨1\,2⟩\,⟨3\,5⟩\,⟨4\,0⟩\,⟨2\,7⟩\right)\,\mathrm{output}\=\mathrm{plot}\right)$