Question

The area of the region bounded by the curves \( x = y^2 - 2 \) and \( x = y \) is

• A. \( \frac{9}{4} \)
• B. 9
• C. \( \frac{9}{2} \)
• D. \( \frac{9}{7} \)

Hint:

When the boundary curves are expressed as \(x = f(y)\), it is often much easier to integrate with respect to \(y\) (Horizontal strips) to avoid splitting the integral into multiple parts.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
The area between two curves is found by integrating the difference between the "right" curve and the "left" curve (if integrating with respect to \(y\)) over the interval defined by their points of intersection.

Step 2: Finding Points of Intersection
Set the two equations equal to each other: \[ y^2 - 2 = y \] \[ y^2 - y - 2 = 0 \implies (y - 2)(y + 1) = 0 \] The curves intersect at \( y = -1 \) and \( y = 2 \).

Step 3: Calculating the Area
In the interval \([-1, 2]\), the line \(x = y\) is to the right of the parabola \(x = y^2 - 2\). \[ \text{Area} = \int_{-1}^{2} [y - (y^2 - 2)] \, dy = \int_{-1}^{2} (y - y^2 + 2) \, dy \] \[ = \left[ \frac{y^2}{2} - \frac{y^3}{3} + 2y \right]_{-1}^{2} \] \[ = \left( \frac{4}{2} - \frac{8}{3} + 4 \right) - \left( \frac{1}{2} - \frac{-1}{3} - 2 \right) \] \[ = \left( 6 - \frac{8}{3} \right) - \left( \frac{1}{2} + \frac{1}{3} - 2 \right) = \frac{10}{3} - \left( -\frac{7}{6} \right) \] \[ = \frac{20 + 7}{6} = \frac{27}{6} = \frac{9}{2} \]

Step 4: Final Answer
The area of the region is \( \frac{9}{2} \).