One of the most beautiful topics after surface integrals is line integrals.
We have to find the line integral:
\(\displaystyle \int_C f(x,y) \ ds\)
We have two ways to solve this problem. One without parametrization.
\(\displaystyle \int_C f(x,y) \ ds = \int_{1}^{3} 2x \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \ dx\)
where \(\displaystyle y = \frac{3}{2}x + \frac{1}{2}\)
I like to call the line integral \(\displaystyle m\) since physically, it means mass.
\(\displaystyle m = \int_C f(x,y) \ ds = \int_{1}^{3} 2x \sqrt{1 + \left(\frac{9}{4}\right)} \ dx = \sqrt{13}\int_{1}^{3} x \ dx\)
Then,
\(\displaystyle m = \frac{\sqrt{13}}{2}\bigg(3^2 - 1^2\bigg) = \textcolor{blue}{4\sqrt{13}}\)
Another way is to solve with parametrization.
\(\displaystyle \int_C f(x,y) \ ds = \int_{0}^{1} 2(1 + 2t) \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \ dt\)
where
\(\displaystyle x(t) = 1 + 2t\)
\(\displaystyle y(t) = 2 + 3t\)
\(\displaystyle m = 2\int_{0}^{1} (1 + 2t) \sqrt{2^2 + 3^2} \ dt = 2\sqrt{13}(t + t^2)\bigg|_{0}^{1} = \textcolor{red}{4\sqrt{13}}\)
