Question

The area enclosed by the curve \(y=-x^2\) and the line \(x+y+2=0\) is:

• A. \(4\) sq units
• B. \(4.5\) sq units
• C. \(5.5\) sq units
• D. \(3.5\) sq units

Hint:

For area between curves: \[ \text{Area}=\int (\text{Upper curve}-\text{Lower curve})\,dx \] Always determine which graph lies above before integrating.

Answer 1

The Correct Option is B

To find the area enclosed by the curve \(y = -x^2\) and the line \(x + y + 2 = 0\), we need to determine the points of intersection and then evaluate the area between them.

1. Find the points of intersection:
   • Substitute \(y = -x^2\) into the line equation \(x + y + 2 = 0\):
   • \(x + (-x^2) + 2 = 0 \rightarrow -x^2 + x + 2 = 0\)
   • This simplifies to the quadratic equation \(-x^2 + x + 2 = 0\).
   • Solving this quadratic equation using factorization or the quadratic formula, we find:
2. Set up the definite integral:
   • The area between the curves from \(x = -1\) to \(x = 2\) can be found by integrating the difference between the functions:
   • This simplifies to \(Area = \int_{-1}^{2} [-x^2 + x + 2] \, dx\).
3. Calculate the definite integral:
   • Integrate term by term: \(\int -x^2 dx = -\frac{x^3}{3}\), \(\int x dx = \frac{x^2}{2}\), and \(\int 2 dx = 2x\).
   • Find the integral from \(-1\) to \(2\):
   • Calculate the definite integral:
     • At \(x = 2\): \(-\frac{(2)^3}{3} + \frac{(2)^2}{2} + 2(2) = -\frac{8}{3} + 2 + 4 = \frac{10}{3}\)
     • At \(x = -1\): \(-\frac{(-1)^3}{3} + \frac{(-1)^2}{2} + 2(-1) = \frac{1}{3} + \frac{1}{2} - 2 = -\frac{7}{6}\)
     • Subtract the results: \(\frac{10}{3} - \left(-\frac{7}{6}\right) = \frac{20}{6} + \frac{7}{6} = \frac{27}{6} = 4.5\)
4. Conclusion:
   • The area enclosed by the curve and the line is \(4.5\) square units.