Problem 5 - Calculus: Early Transcendentals

Question

Consider the function \(f(x)=x^{2}-3 x+2\) on \([0,4] .\) Find the total area between the curve and the \(x\) -axis (measuring all area as positive).

Step-by-Step Solution

Answer

The total area between the curve and the x-axis is \(\frac{28}{3}\).

Step 1: Determine where the function intersects the x-axis

To find the points where the function intersects the x-axis, set the function equal to zero and solve for \(x\):\[ x^2 - 3x + 2 = 0 \]Using the quadratic formula, \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1\), \(b = -3\), \(c = 2\). The discriminant, \(b^2 - 4ac = 1\), is positive.Solving gives:\[ x = \frac{3 \pm 1}{2} \]\[ x = 2 \quad \text{or} \quad x = 1 \]The function intersects the x-axis at points \(x = 1\) and \(x = 2\).

Step 2: Set up the integrals over each interval

The function \(f(x) = x^2 - 3x + 2\) is positive on the interval [0, 1], negative on the interval [1, 2], and positive again on the interval [2, 4]. We compute the absolute values of integrals over each of these intervals to find total area.The integrals are:- From 0 to 1: \( \int_0^1 (x^2 - 3x + 2) \, dx \)- From 1 to 2: \( \int_1^2 |x^2 - 3x + 2| \, dx \)- From 2 to 4: \( \int_2^4 (x^2 - 3x + 2) \, dx \)

Step 3: Calculate each integral

Evaluate each integral:1. From 0 to 1: \[ \int_0^1 (x^2 - 3x + 2) \, dx = \left[ \frac{x^3}{3} - \frac{3x^2}{2} + 2x \right]_0^1 = \left(\frac{1}{3} - \frac{3}{2} + 2\right) = \frac{5}{6} \]2. From 1 to 2 (taking absolute value): \[ \int_1^2 -(x^2 - 3x + 2) \, dx = \int_1^2 (-x^2 + 3x - 2) \, dx = \left[-\frac{x^3}{3} + \frac{3x^2}{2} - 2x \right]_1^2 = \frac{1}{6} \]3. From 2 to 4: \[ \int_2^4 (x^2 - 3x + 2) \, dx = \left[ \frac{x^3}{3} - \frac{3x^2}{2} + 2x \right]_2^4 = \frac{14}{3} \]

Step 4: Sum the absolute values of each integral to find the total area

Add up the absolute values of the computed integrals from each interval to find the total area between the curve and the \(x\)-axis:\[\text{Total Area} = \left|\frac{5}{6}\right| + \left|\frac{1}{6}\right| + \left|\frac{14}{3}\right| = \frac{5}{6} + \frac{1}{6} + \frac{14}{3} = \frac{28}{3}\].

Key Concepts

Quadratic Formula

The quadratic formula is a powerful tool for solving quadratic equations of the form \(ax^2 + bx + c = 0\). It provides the solutions or \'roots\' of the equation using a single equation:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]Here, the coefficients \(a\), \(b\), and \(c\) are from the quadratic expression, specifically:

• \(a\) is the coefficient of \(x^2\),
• \(b\) is the coefficient of \(x\),
• \(c\) is the constant term.

To use the quadratic formula effectively, you must calculate the discriminant, \(b^2 - 4ac\). The discriminant determines the nature of the roots:

• If it\'s positive, there are two distinct real roots.
• If it\'s zero, there\'s exactly one real root (repeated).
• If it\'s negative, the roots are complex.

In our exercise, for \(x^2 - 3x + 2\), \(a = 1\), \(b = -3\), and \(c = 2\). The discriminant is 1, indicating two real intersection points with the \(x\)-axis at \(x = 1\) and \(x = 2\). This method is efficient for finding these crucial intersection points which are important for determining the areas under the curve.

Area Under the Curve

Calculating the area under a curve helps us to understand the spatial region enclosed by the curve and the \(x\)-axis. This is often achieved using definite integrals. For a given function \(f(x)\), the definite integral from \(a\) to \(b\) is expressed as:\[\int_a^b f(x) \, dx\]This operation accumulates the signed area from \(a\) to \(b\). In the context of our exercise, the total area between the curve \(f(x) = x^2 - 3x + 2\) and the \(x\)-axis is of interest over various intervals. To ensure all areas contribute positively to the total, the absolute value of integrals is taken where necessary:

• For intervals where the graph is above the \(x\)-axis, the integral directly gives the area.
• For intervals below the \(x\)-axis, we take the absolute value of the integral to turn the negative area into positive.

This process guarantees the computed total area reflects the true size of the region between the curve and the axis, without concern for the directionality (positional above/below axial) of the curve itself.

Absolute Value of Integrals

In some cases, functions cross the \(x\)-axis, producing regions both above and below it. Normally, integrals provide a negative value for areas below the \(x\)-axis. However, when calculating total area as positive, taking the absolute value is essential. The absolute value of an integral is calculated as:\[A = \int_a^b |f(x)| \, dx\]Where \(A\) is the total positive area obtained, ensuring every part of the function that dips below the \(x\)-axis is treated as if it were above.For \(f(x) = x^2 - 3x + 2\), calculating the absolute value over the interval \([1, 2]\) is central, as \(f(x)\) is negative here. By effectively flipping the curve segment to be above the axis (positively adding to the cumulative area), the full extent of the region's spatial coverage is accurately reflected. This adjustment is crucial for the correct computation of the total area under the curve between specified bounds.

X-Axis Intersection Points

Intersection points with the \(x\)-axis are determined by solving \(f(x) = 0\). These points mark where the graph crosses the axis and are vital for setting partition boundaries in area calculations. For the function \(f(x) = x^2 - 3x + 2\), setting it to zero and using the quadratic formula yielded critical points at \(x = 1\) and \(x = 2\). These values divide the domain into different sections where the behavior of the function concerning the axis changes:

• \([0,1]\) where \(f(x)\) is above the \(x\)-axis,
• \([1,2]\) where \(f(x)\) is below,
• \([2,4]\) where \(f(x)\) again lies above.

Understanding these intersections helps in organizing integral evaluation into practical segments, which are then adjusted to reflect the total positive area. Recognizing and applying the significance of these intersection points enable a comprehensive calculation of the areas surrounded by the function\'s curve and the \(x\)-axis.