Problem 49 - Applied Calculus: For Business, Economics, and the Social and Life Sciences

Question

Use a double integral to find the area of \(R\).\(R\) is the region bounded by \(y=x^{2}-4 x+3\) and the \(x\) axis.

Step-by-Step Solution

Answer

The area of the region is \( \frac{4}{3} \).

Step 1: Identify the Intersection Points

The bounds of the region are determined by the points where the function intersects the x-axis. We need to solve the equation \(y = x^2 - 4x + 3\).Set \(y = 0\): \[x^2 - 4x + 3 = 0.\] Factor the quadratic equation: \[(x - 1)(x - 3) = 0.\] So, the intersection points are \(x = 1\) and \(x = 3\).

Step 2: Set Up the Integral

Now that we know the bounds are from \(x = 1\) to \(x = 3\), we need to set up the double integral to find the area under the curve. The function is \(y = x^2 - 4x + 3\). Thus, the integral is: \[\text{Area} = \int_{1}^{3} \int_{0}^{x^2 - 4x + 3} dy \, dx.\]

Step 3: Integrate with Respect to y

First, integrate with respect to \(y\): \[\text{Area} = \int_{1}^{3} \left[ y \right]_{0}^{x^2 - 4x + 3} dx.\] This simplifies to: \[\text{Area} = \int_{1}^{3} (x^2 - 4x + 3 - 0) \, dx = \int_{1}^{3} (x^2 - 4x + 3) \, dx.\]

Step 4: Integrate with Respect to x

Now, integrate with respect to \(x\): \[\text{Area} = \left[ \frac{x^3}{3} - 2x^2 + 3x \right]_{1}^{3}.\] Evaluate the integral at the bounds: \[ \left( \frac{27}{3} - 2 \cdot 9 + 9 \right) - \left( \frac{1}{3} - 2 \cdot 1 + 3 \right).\]

Step 5: Compute the Definite Integral

Simplify the evaluated integral: \[\text{Area} = (9 - 18 + 9) - \left( \frac{1}{3} - 2 + 3 \right) = 0 - \left( \frac{1}{3} + 1 \right).\] This simplifies further: \[\text{Area} = - \left( \frac{1}{3} + 1 \right) = - \left( \frac{4}{3} \right).\]

Step 6: Final Area

Since the integral represents the area and it cannot be negative, take the absolute value: \[\text{Area} = \frac{4}{3}.\]

Key Concepts

Intersection Points

To start calculating the area, we need to find the points where the function intersects the x-axis. This gives us the boundary for our region of interest.
The function we have is: \(y = x^2 - 4x + 3\).
Set \(y = 0\): \[x^2 - 4x + 3 = 0.\]
This is a quadratic equation, and we solve it by factoring: \[(x - 1)(x - 3) = 0.\]
From this, we get the intersection points as \(x = 1\) and \(x = 3.\) These are the \'roots\' of the quadratic equation and serve as the limits for our integrals later on.

Integral Bounds

With the intersection points identified, we now set the bounds for our double integral.
The region \(R\) is bounded by \(x = 1\) and \(x = 3\). Inside this region, the upper limit is defined by the curve \((y = x^2 - 4x + 3)\) and the lower limit is the x-axis, which is \(y = 0\).
Therefore, setting up the double integral for the area, we have:

• The outer integral (with respect to \(x\)) goes from \(1\) to \(3\).
• The inner integral (with respect to \(y\)) ranges from \(0\) to \((x^2 - 4x + 3).\)

The integral is: \[ \text{Area} = \int_{1}^{3} \int_{0}^{x^2 - 4x + 3} dy \, dx.\]
This sets up our complete integral bounds for calculating the area.

Definite Integral

A definite integral calculates the net area under a curve within a specific interval.
Here, we need to find the area between \(x = 1\) and \(x = 3\) under the curve \((y = x^2 - 4x + 3)\).
Once our integral bounds are set, we translate this into a solvable form: \[ \text{Area} = \int_{1}^{3} \int_{0}^{x^2 - 4x + 3} dy \, dx.\]
The inner integral integrates with respect to \(y\). Since our function in terms of \(y\) is \(x^2 - 4x + 3\), integrating within those bounds:
\[ \text{Area} = \int_{1}^{3} \left[ y \right]_{0}^{x^2 - 4x + 3} dx.\]
This evaluates to: \[ \text{Area} = \int_{1}^{3} (x^2 - 4x + 3 - 0) \, dx = \int_{1}^{3} (x^2 - 4x + 3) \, dx.\]

Integration Steps

Finally, we perform the integration steps:

• First, integrate with respect to \(y\).

\[ \text{Area} = \int_{1}^{3} \left[ y \right]_{0}^{x^2 - 4x + 3} dx.\]
\[ \text{Area} = \int_{1}^{3} (x^2 - 4x + 3 - 0) \, dx = \int_{1}^{3} (x^2 - 4x + 3) \, dx.\]

• After integrating with respect to \(x\), we get: \ \text{Area} = \left[ \frac{x^3}{3} - 2x^2 + 3x\right]_{1}^{3}.

Evaluate the integral at the bounds: \[ \left( \frac{27}{3} - 18 + 9 \right) - \left( \frac{1}{3} - 2 + 3 \right). \]
\[ \text{Area} = (9 - 18 + 9) - \left( \frac{1}{3} + 1 \right). \]

• Simplify the evaluated integral: \[ \text{Area} = - ( \frac{1}{3} + 1 ) = - \frac{4}{3}. \] Because the area cannot be negative, we take the absolute value: \[ \text{Area} = \frac{4}{3}.\]

Through these steps, we systematically solve the double integral to find the bounded area.