Given the parametric equations $x=t^2-2t$ and $y=t$ for $t\in [0,2]$, what is the area enclosed by the curve and the y-axis?

• A. $\frac{8}{3}$
• B. $\frac{16}{3}$
• C. $4$
• D. $2$

Answer: (A)

Verified step by step guidance

1. Step 1: Recall that the area enclosed by a parametric curve and the y-axis can be calculated using the formula \( A = \int y \frac{dx}{dt} dt \), where \( y \) and \( \frac{dx}{dt} \) are derived from the parametric equations.
2. Step 2: Compute \( \frac{dx}{dt} \) by differentiating \( x = t^2 - 2t \) with respect to \( t \). This gives \( \frac{dx}{dt} = 2t - 2 \).
3. Step 3: Substitute \( y = t \) and \( \frac{dx}{dt} = 2t - 2 \) into the area formula \( A = \int y \frac{dx}{dt} dt \). The integral becomes \( A = \int_{0}^{2} t (2t - 2) dt \).
4. Step 4: Expand the integrand \( t (2t - 2) \) to simplify the expression. This gives \( A = \int_{0}^{2} (2t^2 - 2t) dt \).
5. Step 5: Evaluate the integral \( \int_{0}^{2} (2t^2 - 2t) dt \) by integrating term by term. First, integrate \( 2t^2 \) to get \( \frac{2t^3}{3} \), and integrate \( -2t \) to get \( -t^2 \). Then, apply the limits of integration from \( t = 0 \) to \( t = 2 \).