Problem 59 - University Calculus: Early Transcendentals

Question

Find the total area between the region and the \(x\) -axis. $$y=x^{3}-3 x^{2}+2 x, \quad 0 \leq x \leq 2$$

Step-by-Step Solution

Answer

The total area is \(\frac{1}{2}\).

Step 1: Understand the Problem

We are asked to find the total area between the curve represented by the function \( y = x^3 - 3x^2 + 2x \) and the \(x\)-axis over the interval \(0 \leq x \leq 2\). This involves identifying where the curve is above and below the \(x\)-axis.

Step 2: Find Points of Intersection with the x-axis

The intersections with the \(x\)-axis occur where \( y = 0 \). Solve the equation \( x^3 - 3x^2 + 2x = 0 \). Factor out \(x\):\[ x(x^2 - 3x + 2) = 0 \]Then factor \(x^2 - 3x + 2\):\[ x(x-1)(x-2) = 0 \]The solutions are \( x = 0, 1, 2 \). Here are our points of intersection.

Step 3: Determine Intervals to Consider

The points of intersection divide the interval \( [0, 2] \) into subintervals \([0, 1]\), \([1, 2]\). We need to find the area in each subinterval where the curve is either above or below the \(x\)-axis to correctly determine the total enclosed area.

Step 4: Calculate Area in the Interval [0, 1]

In the interval \([0, 1]\), the function is above the \(x\)-axis. The area can be found using the definite integral from 0 to 1:\[ \int_0^1 (x^3 - 3x^2 + 2x) \, dx \].Calculate this:1. Find the antiderivative: \( \frac{x^4}{4} - x^3 + x^2 \).2. Evaluate from 0 to 1: \[ \left( \frac{1}{4} - 1 + 1 \right) - \left( 0 \right) = \frac{1}{4} \].

Step 5: Calculate Area in the Interval [1, 2]

In the interval \([1, 2]\), the function is below the \(x\)-axis. The area is:\[ \int_1^2 -(x^3 - 3x^2 + 2x) \, dx \].Calculate this integral:1. The antiderivative is \( \frac{x^4}{4} - x^3 + x^2 \).2. Evaluate from 1 to 2: \[ \left( \frac{16}{4} - 8 + 4 \right) - \left( \frac{1}{4} - 1 + 1 \right) = 0 - \frac{1}{4} = -\frac{1}{4} \].Since we're finding the total area, take the absolute value, resulting in \(\frac{1}{4}\).

Step 6: Add the Areas from Each Interval

To find the total area, add the absolute values of the areas calculated from each interval:\[ \text{Total Area} = \left| \frac{1}{4} \right| + \left| -\frac{1}{4} \right| = \frac{1}{4} + \frac{1}{4} = \frac{1}{2} \].

Key Concepts

Definite Integrals

Definite integrals are fundamental in calculus for finding the area under curves and between curves. When solving problems involving definite integrals, we often look at the integral of a function over a specific interval to determine these areas. The given problem asks us to find the total area between the curve of the function \( y = x^3 - 3x^2 + 2x \) and the \( x \)-axis over the interval \( [0, 2] \).
This involves computing separate integrals over subintervals where the function changes its position relative to the \( x \)-axis.

To apply definite integrals:

• First, identify the points of intersection of the curve with the \( x \)-axis, which occur where the function value is zero.
• Then, calculate the definite integral over each subinterval that forms between these points of intersection.

It's important to also recognize if the curve is above or below the \( x \)-axis in each of these subintervals, as it affects the way we sum the areas.

Area Between Curves

The problem we're dealing with involves finding the area between a curve and the \( x \)-axis. This is a specific instance of a more generalized concept in calculus known as 'finding the area between curves'.

To find the area between a curve and the \( x \)-axis:

• Identify the intervals over which the function is above or below the \( x \)-axis.
• If the function is above the axis, compute the integral of the function over the interval.
• If it's below the axis, compute the integral of the negative of the function to get a positive area value.

This approach ensures that all calculated areas are positive, which is essential when they are summed to determine the total area. In our problem, we handle each interval separately to find the regions under the curve, which are then combined to get the full area between the curve and the \( x \)-axis over the specified range.

Antiderivatives

Antiderivatives are the foundation for solving integral calculus problems. They represent the reverse process of differentiating a function. To find the area under a curve, we must first determine its antiderivative.

In our exercise, the original function given is \( y = x^3 - 3x^2 + 2x \). To solve the integrals:\

• First, we find the antiderivative, which involves integrating each term of the function separately.
• The antiderivative turns out to be \( \frac{x^4}{4} - x^3 + x^2 \).

Once we have this antiderivative, we calculate the definite integrals to get the area under or above the \( x \)-axis. Calculating this involves substituting the limits of the interval into the antiderivative and finding the difference, giving us the precise areas needed for the total calculation. Understanding and finding antiderivatives is a crucial skill in both engineering and physical sciences, where such calculations are routinely required.