# 008. Parametric Area

#### Given a curve such that $x = t^2 +1$ and $y = t^2 + t - 2$ for all $t \in \mathbb{R}$, calculate the area betwen this curve and $x$-axis.

To find the intersection points between this curve and $x$-axis, set $y = 0$. So, the intersection points are $x = 5, 2$. Besides, $x = t^2 + 1$ is minimized to $x = 1$ when $t = 0$. \begin{aligned} y = t^2 + t - 2 = (t + 2)(t - 1) = 0 \implies t = -2, 1 \end{aligned}

In general, the parametric area can be calculated from the following formula when $x = g(t)$, $y = f(t)$, $a = g(\alpha)$, and $b = g(\beta)$. Noting that $dx = g'(t)dt$, \begin{aligned} \int_{a}^{b} y dx = \int_{\alpha}^{\beta} y g'(t)dt = \int_{\alpha}^{\beta} f(t) g'(t)dt \end{aligned}

Therefore, the area is \begin{aligned} \text{Area} &= \left \vert \int_{0}^{-2} f(t) g'(t)dt \right \vert - \left \vert \int_{0}^{1} f(t) g'(t)dt \right \vert = -\int_{0}^{-2} f(t) g'(t)dt + \int_{0}^{1} f(t) g'(t)dt \\\\ &= \int_{-2}^{0} f(t) g'(t)dt + \int_{0}^{1} f(t) g'(t)dt = \int_{-2}^{1} f(t) g'(t)dt \\\\ &= \int_{-2}^{1} (t^2 + t - 2)(2t) dt = \int_{-2}^{1} 2t^3 + 2t^2 - 4t dt \\\\ &= \left[ \cfrac{t^4}{2} + \cfrac{2t^3}{3} - 2t^2\right]_{-2}^{1} = \left( \cfrac{1}{2} + \cfrac{2}{3} - 2 \right) - \left( 8 - \cfrac{16}{3} - 8 \right) = \cfrac{9}{2} \end{aligned}