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

Concept: The area enclosed between two curves is calculated using definite integration: \[ \text{Area} = \int_a^b (\text{Upper curve}-\text{Lower curve})\,dx \] The first step is always to find the points of intersection of the two curves.

Step 1: Write the equations in convenient form. The parabola is: \[ y=-x^2 \] The line is: \[ x+y+2=0 \] which can be written as: \[ y=-x-2 \]

Step 2: Find the points of intersection. At the points of intersection: \[ -x^2=-x-2 \] Bring all terms to one side: \[ x^2-x-2=0 \] Factorize: \[ (x-2)(x+1)=0 \] Thus, \[ x=2 \quad \text{or} \quad x=-1 \] Hence the curves intersect at: \[ x=-1 \quad \text{and} \quad x=2 \]

Step 3: Determine the upper and lower curve. Take a point between \(-1\) and \(2\), say \(x=0\). For the parabola: \[ y=-0^2=0 \] For the line: \[ y=-0-2=-2 \] Since \(0>-2\), the parabola lies above the line. Thus, \[ \text{Area} = \int_{-1}^{2} \left[ (-x^2)-(-x-2) \right]dx \] \[ = \int_{-1}^{2} (-x^2+x+2)\,dx \]

Step 4: Evaluate the definite integral. Integrating term-by-term: \[ \int(-x^2+x+2)\,dx = -\frac{x^3}{3} +\frac{x^2}{2} +2x \] Now apply limits: \[ \text{Area} = \left[ -\frac{x^3}{3} +\frac{x^2}{2} +2x \right]_{-1}^{2} \] At \(x=2\): \[ -\frac{8}{3} +\frac{4}{2} +4 = -\frac83+2+4 = \frac{10}{3} \] At \(x=-1\): \[ -\left(\frac{-1}{3}\right) +\frac12 -2 = \frac13+\frac12-2 \] \[ = \frac{2+3-12}{6} = -\frac76 \] Thus, \[ \text{Area} = \frac{10}{3}-\left(-\frac76\right) \] \[ = \frac{20}{6}+\frac76 = \frac{27}{6} = \frac92 \] \[ =4.5 \] Hence, \[ \boxed{4.5\ \text{sq units}} \]