Question

The area bounded by the curve \(x = 2 - y - y^2\) and the Y-axis is

• A. \(7/6\)
• B. \(13/2\)
• C. \(9/2\)
• D. \(27/2\)

Hint:

When the curve is given as \(x=f(y)\), it is usually best to integrate with respect to \(y\).

Answer 1

The Correct Option is C

Concept:
The y-axis is given by: \[ x=0 \] So the required area is the area between: \[ x=2-y-y^2 \quad \text{and} \quad x=0 \] with respect to \(y\). ip

Step 1: Find the points of intersection with the y-axis.
Put \[ x=0 \] in the curve: \[ 2-y-y^2=0 \] \[ y^2+y-2=0 \] \[ (y+2)(y-1)=0 \] So the limits are: \[ y=-2,\ 1 \] ip

Step 2: Set up the area integral.
Area: \[ A=\int_{-2}^{1}(2-y-y^2)\,dy \] ip

Step 3: Evaluate the integral.
\[ \int (2-y-y^2)\,dy = 2y-\frac{y^2}{2}-\frac{y^3}{3} \] So, \[ A=\left[2y-\frac{y^2}{2}-\frac{y^3}{3}\right]_{-2}^{1} \] At \(y=1\): \[ 2-\frac12-\frac13=\frac{7}{6} \] At \(y=-2\): \[ -4-\frac{4}{2}-\frac{-8}{3} =-4-2+\frac{8}{3} =-\frac{10}{3} \] Therefore, \[ A=\frac{7}{6}-\left(-\frac{10}{3}\right) =\frac{7}{6}+\frac{20}{6} =\frac{27}{6} =\frac{9}{2} \] ip Hence, the correct answer is:
\[ \boxed{(C)\ \frac{9}{2}} \]