3.7 Q-6

Question Statement

Determine the area bounded by the parabola $y = x^2 + 2x - 3$ and the $x$-axis.

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Background and Explanation

In this problem, we need to find the area enclosed between the given parabola and the $x$-axis. To do so, we must first identify the points where the curve intersects the $x$-axis, i.e., where $y = 0$. Afterward, we can compute the definite integral of the function between these points to calculate the area under the curve. The integration will account for the portion of the curve that lies below the $x$-axis.

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Solution

Step 1: Set the equation equal to zero

To find the $x$-intercepts, we set $y = 0$ in the equation of the parabola:

$$x^2 + 2x - 3 = 0$$

Step 2: Solve the quadratic equation

We solve the quadratic equation using factoring:

$$x^2 + 2x - 3 = (x + 3)(x - 1) = 0$$

This gives the solutions:

$$x = -3 \quad \text{or} \quad x = 1$$

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

Step 3: Set up the integral

Since the parabola is below the $x$-axis between $x = -3$ and $x = 1$, the area between the curve and the $x$-axis can be computed by integrating the function $y = x^2 + 2x - 3$ from $x = -3$ to $x = 1$:

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

Step 4: Compute the integral

We now compute the integral of each term in $x^2 + 2x - 3$:

$$\int x^2 , dx = \frac{x^3}{3}, \quad \int 2x , dx = x^2, \quad \int -3 , dx = -3x$$

Substitute these into the integral:

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

Step 5: Evaluate the definite integral

Now, evaluate the expression at the limits $x = 1$ and $x = -3$:

$$A = \left( \frac{1^3}{3} + 1^2 - 3(1) \right) - \left( \frac{(-3)^3}{3} + (-3)^2 - 3(-3) \right)$$

Simplifying both terms:

For $x = 1$:

$$\frac{1^3}{3} + 1^2 - 3(1) = \frac{1}{3} + 1 - 3 = \frac{1}{3} - 2 = -\frac{5}{3}$$

For $x = -3$:

$$\frac{(-3)^3}{3} + (-3)^2 - 3(-3) = \frac{-27}{3} + 9 + 9 = -9 + 18 = 9$$

Step 6: Final simplification

Now, subtract the two results:

$$A = \left( -\frac{5}{3} \right) - 9 = -\frac{5}{3} - \frac{27}{3} = \frac{-28}{3}$$

Since the area is always positive, we take the absolute value:

$$A = \frac{28}{3} \text{ square units}$$

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Key Formulas or Methods Used

• Definite Integral: The area between a curve and the $x$-axis is given by the definite integral:

$$A = \int_a^b f(x) , dx$$

• Power Rule of Integration:

$$\int x^n , dx = \frac{x^{n+1}}{n+1}$$

• Factoring a Quadratic Equation: The roots of a quadratic equation $ax^2 + bx + c = 0$ can be found by factoring the equation into two binomials.

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Summary of Steps

1. Set the equation of the parabola equal to zero to find the $x$-intercepts: $x^2 + 2x - 3 = 0 \quad \Rightarrow \quad x = -3, x = 1$
2. Set up the integral to compute the area: $A = \int_{-3}^{1} (x^2 + 2x - 3) , dx$
3. Compute the integral: $A = \left[ \frac{x^3}{3} + x^2 - 3x \right]_{-3}^{1}$
4. Evaluate the integral at the limits: $A = -\frac{5}{3} - 9 = \frac{-28}{3}$
5. Take the absolute value for the final area: $A = \frac{28}{3} , \text{square units}$