I'm integrating some parametric functions and I came across this one and it's confusing since it fundamentally doesn't make sense
$$x = t^5 - 4t^3$$ $$y = t^2$$
How would I approach this problem, if I were to, say integrate from t = -3 to t = 3?
This integration doesn't make sense since integration is fundamentally "area under a curve" right? With this curve, the curve overlaps so I'm not sure how to approach this.
Your path must be closed in order to form an area. $[-3,3]$ will not do.
$[-2,2]$ will work.
That is: $x(-2) = x(2)$ and $y(-2) = y(2)$
Area: $\int y \ dx = -\int x\ dy$
You can solve either one... but just do demonstrate that the two are equal, I will do both.
$\int_{-2}^2 t^2(5t^4 -12t^2) \ dt = -\int_{-2}^2 (t^5 -4t^3)(2t) \ dt\\ \int_{-2}^2 5t^6 -12t^4 \ dt = -\int_{-2}^2 2t^6 -8t^4 \ dt\\ 2(\frac {5}{7} 2^7 - \frac {12}{5} 2^5) = -2(\frac {2}{7}2^7 - \frac {8}{5} 2^5)\\ 2(\frac {512}{35}) = - 2(-\frac {512}{35}) $