Next, we need to parameterize the curve.

$$\vec r\left( t \right) = \left\langle {t,1 - {t^3}} \right\rangle \hspace{0.25in} - 1 \le t \le 2$$ Show Step 3

Now we need to evaluate the line integral. Be careful with this type of line integral. In this case we have both a $dx$ and a $dy$ in the integrand. Recall that this is just a simplified notation for,

$$\displaystyle \int\limits_{C}{{2y\,dx + \left( {1 - x} \right)\,dy}} = \int\limits_{C}{{2y\,dx}} + \int\limits_{C}{{1 - x\,dy}}$$

Then all we need to do is recall that $dx = x'\,dt$ and $dy = y'\,dt$ when we convert the line integral into a “standard” integral.

So, let’s evaluate the line integral. Just remember to “plug in” the parameterization into the integrand (i.e. replace the $x$ and $y$ in the integrand with the $x$ and $y$ components of the parameterization) and to convert the differentials properly.

Here is the line integral.

$$\begin{align*}\int\limits_{C}{{2y\,dx + \left( {1 - x} \right)\,dy}} &= \int\limits_{C}{{2y\,dx}} + \int\limits_{C}{{1 - x\,dy}}\\ & = \int_{{ - 1}}^{2}{{2\left( {1 - {t^3}} \right)\,\,\left( 1 \right)\,dt}} + \int_{{ - 1}}^{2}{{\left( {1 - t} \right)\,\,\left( { - 3{t^2}} \right)\,dt}}\\ & = \int_{{ - 1}}^{2}{{2\left( {1 - {t^3}} \right)\,dt}} - 3\int_{{ - 1}}^{2}{{{t^2} - {t^3}\,dt}}\\ & = \int_{{ - 1}}^{2}{{{t^3} - 3{t^2} + 2\,dt}}\\ & = \left. {\left[ {\frac{1}{4}{t^4} - {t^3} + 2t} \right]} \right|_{ - 1}^2 = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{3}{4}}}\end{align*}$$

Note that, in this case, we combined the two integrals into a single integral prior to actually evaluating the integral. This doesn’t need to be done but can, on occasion, simplify the integrand and hence the evaluation of the integral.