${\int }_C\mathbf{F}·\mathbf{dr}$

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

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

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

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

$\=17\/12$

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 }$${\∫}_0^1\left(t^3\+2\left(t^2\+t\right)t\right)\ⅆt$$\=$$\frac{17}{12}$

Define the path parametrically, as a position vector

$\mathbf{R}\=⟨t\,t^2⟩$$\stackrel{\text{assign}}{\to }$$$

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{4t^2\+1}$$\stackrel{\text{assign to a name}}{\to }$$\mathrm{\ρ}$$$

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

$\mathbf{F}·\mathbf{T} \mathrm{\ρ}$

$\left(\frac{t^3}{\sqrt{4t^2\+1}}\+\frac{2\left(t^2\+t\right)t}{\sqrt{4t^2\+1}}\right)\sqrt{4t^2\+1}$

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

$t^2\left(2\+3t\right)$

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

${\∫}_0^1{t}^2\left(2\+3t\right)\ⅆt$

$\=$

$\frac{17}{12}$

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

$\mathrm{LineInt}\left(\mathbf{F}\,\mathrm{Path}\left(⟨t\,t^2⟩\,t\=0..1\right)\,\mathrm{output}\=\mathrm{integral}\right)\=\mathrm{LineInt}\left(\mathbf{F}\,\mathrm{Path}\left(⟨t\,t^2⟩\,t\=0..1\right)\right)$

${\∫}_0^1\left(t^3\+2\left(t^2\+t\right)t\right)\ⅆt\=\frac{17}{12}$

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

$\mathrm{LineInt}\left(\mathbf{F}\,\mathrm{Path}\left(⟨t\,t^2⟩\,t\=0..1\right)\,\mathrm{output}\=\mathrm{plot}\right)$

Define the path parametrically, as the position vector R

$\mathbf{R}≔⟨t\,t^2⟩\:$

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

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

$\sqrt{4t^2\+1}$

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

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

${\∫}_0^1\left(\frac{t^3}{\sqrt{4t^2\+1}}\+\frac{2\left(t^2\+t\right)t}{\sqrt{4t^2\+1}}\right)\sqrt{4t^2\+1}\ⅆt\=\frac{17}{12}$