Calculate the area bounded by the curves y = 3 - 2x - x²

Calculate the area bounded by the curves $y = 3 - 2x - x^2$ and $y = 0$.

We will evaluate the area between the parabola and the $x$-axis.

Step 1: Find intersection points

The curves intersect where:

\[3 - 2x - x^2 = 0\]

Multiplying by $-1$:

\[x^2 + 2x - 3 = 0\]

Factoring:

\[(x + 3)(x - 1) = 0\]

Therefore: $x = -3$ or $x = 1$

The parabola intersects the $x$-axis at $x = -3$ and $x = 1$.

Step 2: Determine which curve is above

The parabola $y = 3 - 2x - x^2$ can be rewritten as:

\[y = -(x^2 + 2x - 3) = -(x + 3)(x - 1)\]

For $x \in (-3, 1)$:

• $(x + 3) > 0$ (positive)
• $(x - 1) < 0$ (negative)
• $(x + 3)(x - 1) < 0$ (negative)
• $-(x + 3)(x - 1) > 0$ (positive)

So the parabola is above the $x$-axis on $(-3, 1)$. ✓

Let's verify at $x = 0$:

\[y = 3 - 0 - 0 = 3 > 0 \quad \checkmark\]

Step 3: Set up the integral

\[A = \int_{-3}^{1} (3 - 2x - x^2) \, dx\]

Step 4: Calculate the antiderivative

\[\int (3 - 2x - x^2) \, dx = 3x - x^2 - \frac{x^3}{3} + C\]

Step 5: Evaluate the definite integral

\[A = \left[3x - x^2 - \frac{x^3}{3}\right]_{-3}^{1}\]

At $x = 1$:

\[3(1) - 1^2 - \frac{1^3}{3} = 3 - 1 - \frac{1}{3} = 2 - \frac{1}{3} = \frac{6 - 1}{3} = \frac{5}{3}\]

At $x = -3$:

\[3(-3) - (-3)^2 - \frac{(-3)^3}{3} = -9 - 9 - \frac{-27}{3}\]

\[= -9 - 9 + 9 = -9\]

Therefore:

\[A = \frac{5}{3} - (-9) = \frac{5}{3} + 9 = \frac{5 + 27}{3} = \frac{32}{3}\]

\[\boxed{A = \frac{32}{3} \text{ square units}}\]

Step 6: Numerical value

\[A = \frac{32}{3} \approx 10.667 \text{ square units}\]

Verification

Let's verify the calculation at $x = -3$:

\[3(-3) - 9 + 9 = -9 - 9 + 9 = -9 \quad \checkmark\]

And:

\[\frac{5}{3} + 9 = \frac{5 + 27}{3} = \frac{32}{3} \quad \checkmark\]

Alternative: Complete the square

We can also write the parabola in vertex form:

\[y = -(x^2 + 2x - 3) = -(x^2 + 2x + 1 - 1 - 3)\]

\[= -((x + 1)^2 - 4) = -(x + 1)^2 + 4 = 4 - (x + 1)^2\]

This shows the vertex is at $(-1, 4)$, opening downward.

Summary

The area bounded by the parabola $y = 3 - 2x - x^2$ and the $x$-axis is:

\[\boxed{A = \frac{32}{3} \approx 10.667 \text{ square units}}\]

Key steps:

1. Find intersections: $x^2 + 2x - 3 = 0 \implies x = -3, 1$
2. Verify parabola is above $x$-axis on $(-3, 1)$
3. Set up integral: $A = \int_{-3}^{1} (3 - 2x - x^2) \, dx$
4. Find antiderivative: $3x - x^2 - \frac{x^3}{3}$
5. Evaluate: $\frac{5}{3} - (-9) = \frac{32}{3}$

Key insight:

The parabola opens downward and has roots at $x = -3$ and $x = 1$, with vertex at $x = -1$ where $y = 4$.