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

• A. $2$
• B. $1$
• C. $\frac{4}{3}$
• D. $\frac{8}{3}$

Answer: (C)

Verified step by step guidance

1. Step 1: Recall the formula for the area enclosed by a parametric curve and the y-axis. The area is given by \( A = \int y \cdot \frac{dx}{dt} \, dt \), where \( \frac{dx}{dt} \) is the derivative of \( x \) with respect to \( t \).
2. Step 2: Compute \( \frac{dx}{dt} \) from the given parametric equation \( x = t^2 - 2t \). Differentiate \( x \) with respect to \( t \): \( \frac{dx}{dt} = 2t - 2 \).
3. Step 3: Substitute \( y = t \) and \( \frac{dx}{dt} = 2t - 2 \) into the area formula: \( A = \int_{0}^{2} t \cdot (2t - 2) \, dt \).
4. Step 4: Expand the integrand \( t \cdot (2t - 2) \) to simplify the integral: \( t \cdot (2t - 2) = 2t^2 - 2t \). The integral becomes \( A = \int_{0}^{2} (2t^2 - 2t) \, dt \).
5. Step 5: Break the integral into two parts: \( A = \int_{0}^{2} 2t^2 \, dt - \int_{0}^{2} 2t \, dt \). Compute each integral separately to find the total area.