Given the parametric equations $x=t^2-3t$ and $y=t$, for $t\in [0,3]$, find the area enclosed by the curve and the y-axis.

• A. $\frac{9}{2}$
• B. $\frac{27}{2}$
• C. $3$
• D. $6$

Answer: (A)

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 \( x = t^2 - 3t \). Differentiate \( x \) with respect to \( t \): \( \frac{dx}{dt} = 2t - 3 \).
3. Step 3: Substitute \( y = t \) and \( \frac{dx}{dt} = 2t - 3 \) into the area formula \( A = \int y \cdot \frac{dx}{dt} \, dt \). This gives \( A = \int_{0}^{3} t \cdot (2t - 3) \, dt \).
4. Step 4: Expand the integrand \( t \cdot (2t - 3) \) to simplify the integral. This becomes \( A = \int_{0}^{3} (2t^2 - 3t) \, dt \).
5. Step 5: Compute the integral \( \int_{0}^{3} (2t^2 - 3t) \, dt \) by integrating each term separately. Use the power rule for integration: \( \int t^n \, dt = \frac{t^{n+1}}{n+1} \). After integrating, evaluate the result at the bounds \( t = 0 \) and \( t = 3 \).