Question

The area of the region bounded by the parabola \( x = -y^2 \) and the line \( y = x + 2 \) equals

• A. \( \frac{3}{2} \)
• B. \( \frac{7}{2} \)
• C. \( \frac{9}{2} \)
• D. 9

Hint:

- When finding the area between curves, always subtract the left function from the right one.
- Set up your integral with correct limits of integration based on the points of intersection.
- The integral gives the signed area; always take the absolute value if the result is negative.

Answer 1

The Correct Option is C

1) Finding the points of intersection:
We are given the equations \( x = -y^2 \) and \( y = x + 2 \). First, substitute \( x = -y^2 \) into the second equation to find the points of intersection.
\[ y = (-y^2) + 2 \] \[ y^2 + y - 2 = 0 \] Factoring the quadratic equation: \[ (y + 2)(y - 1) = 0 \] Thus, the points of intersection are \( y = -2 \) and \( y = 1 \).
2) Setting up the integral for the area:
To find the area, we need to calculate the integral of the difference between the functions over the interval \( y = -2 \) to \( y = 1 \). Since \( x = -y^2 \) is the left boundary and \( x = y - 2 \) (derived from the second equation) is the right boundary, the area is given by: \[ \text{Area} = \int_{-2}^{1} \left[ (y - 2) - (-y^2) \right] dy \] \[ \text{Area} = \int_{-2}^{1} \left( y - 2 + y^2 \right) dy \] 3) Computing the integral:
Now, integrate term by term: \[ \int_{-2}^{1} y \, dy = \left[ \frac{y^2}{2} \right]_{-2}^{1} = \frac{1}{2} - \frac{4}{2} = -\frac{3}{2} \] \[ \int_{-2}^{1} -2 \, dy = -2 \left[ y \right]_{-2}^{1} = -2(1 - (-2)) = -2(3) = -6 \] \[ \int_{-2}^{1} y^2 \, dy = \left[ \frac{y^3}{3} \right]_{-2}^{1} = \frac{1^3}{3} - \frac{(-2)^3}{3} = \frac{1}{3} - \frac{-8}{3} = \frac{9}{3} = 3 \] 4) Summing up the results:
Adding up all the terms, we get the total area: \[ \text{Area} = -\frac{3}{2} - 6 + 3 = -\frac{3}{2} - 3 = -\frac{9}{2} \] Since we are taking the absolute value of the area, the final answer is \( \frac{9}{2} \).