Next, we need to parameterize the curve.

$$\vec r\left( t \right) = \left\langle {t,{{\bf{e}}^{2t}}} \right\rangle \hspace{0.25in}0 \le t \le 2$$ Show Step 3

Now we need to evaluate the line integral. Be careful with this type line integral. Note that the differential, in this case, is $dy$ and not $ds$ as they were in the previous section.

All we need to do is recall that $dy = y'\,dt$ 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 differential properly.

Here is the line integral.

$$\begin{align*}\int\limits_{C}{{\sqrt {1 + y} \,dy}} &= \int_{0}^{2}{{\sqrt {1 + {{\bf{e}}^{2t}}} \,\,\left( {2{{\bf{e}}^{2t}}} \right)\,dt}}\\ & = \left. {\left[ {\frac{2}{3}{{\left( {1 + {{\bf{e}}^{2t}}} \right)}^{\frac{3}{2}}}} \right]} \right|_0^2 = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{2}{3}\left[ {{{\left( {1 + {{\bf{e}}^4}} \right)}^{\frac{3}{2}}} - {2^{\frac{3}{2}}}} \right] = 274.4897}}\end{align*}$$

Note that, in this case, the integral ended up being a simple substitution.