Problem 15 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 11-30, sketch the region bounded by the graphs of the given equations, show a typical slice, approximate its area, set up an integral, and calculate the area of the region. Make an estimate of the area to confirm your answer: \(y=\frac{1}{4}\left(x^{2}-7\right), y=0\), between \(x=0\) and \(x=2\)

Step-by-Step Solution

Answer

The area is \( \frac{17}{6} \) square units.

Step 1: Understand the Problem

We need to find the area between the curves defined by the equations \( y = \frac{1}{4}(x^2 - 7) \) and \( y = 0 \) (the x-axis) from \( x = 0 \) to \( x = 2 \). The area will be the region above the x-axis and below the curve \( y = \frac{1}{4}(x^2 - 7) \).

Step 2: Sketch the Graph

Draw the parabolic curve \( y = \frac{1}{4}(x^2 - 7) \). It opens upwards as the coefficient of \( x^2 \) is positive. The x-axis \( y = 0 \) is a horizontal line. The parabola intersects the x-axis at the points where \( \frac{1}{4}(x^2 - 7) = 0 \), i.e., \( x^2 = 7 \), so \( x = \sqrt{7} \approx 2.65 \). The region we need is bounded by x from 0 to 2.

Step 3: Determine a Typical Slice

A typical slice is a vertical line at a point \( x = a \) between \( x = 0 \) and \( x = 2 \). The height of this slice is determined by \( y = \frac{1}{4}(x^2 - 7) \) since it is above the x-axis.

Step 4: Set Up the Integral

Calculate the area of the region by integrating the height of each slice with respect to x, from 0 to 2. The integral is: \[ A = \int_{0}^{2} \left( \frac{1}{4}(x^2 - 7) \right) \, dx \].

Step 5: Calculate the Definite Integral

Compute the integral to find the area:1. Integrate \( \frac{1}{4}x^2 \) to get \( \frac{1}{12}x^3 \).2. Integrate \( -\frac{7}{4} \) to get \( -\frac{7}{4}x \).3. Evaluate from 0 to 2: \[ \left[ \frac{1}{12}x^3 - \frac{7}{4}x \right]_{0}^{2} = \left( \frac{1}{12}(2)^3 - \frac{7}{4}(2) \right) - \left( \frac{1}{12}(0)^3 - \frac{7}{4}(0) \right) \].

Step 6: Simplify and Evaluate

Let's simplify the expression after substitution:- \( \frac{1}{12}(2)^3 = \frac{8}{12} = \frac{2}{3} \).- \( \frac{7}{4}(2) = \frac{14}{4} = \frac{7}{2} \).Thus, the area is \[ A = \left( \frac{2}{3} - \frac{7}{2} \right) = \frac{4}{6} - \frac{21}{6} = -\frac{17}{6} \].Since area cannot be negative, the absolute value is taken, and the answer is \( \frac{17}{6} \) square units.

Key Concepts

Definite Integrals

Definite integrals are an essential tool in calculus used for calculating the area under a curve. They are not just about finding areas but also encompass accumulation of quantities over an interval. The integral's limits, known as the upper and lower bounds, determine the section of the curve or line segment that we are considering.

When solving area problems like finding the area between curves, definite integrals provide a streamlined approach. You set up an integral with the function representing the top curve minus the function of the bottom curve, between two x-values. This integral gives you the net area contained within that region.

• The lower limit of the integral is where we start measuring, and the upper limit is where we stop.
• The integrand is the function or expression inside the integral, effectively giving us the height of the curve at different x-values.
• Calculating a definite integral involves finding the antiderivative of the integrand and substituting the limits to find the net area.

Integrals not only help in mathematical exercises but also have practical applications in fields like physics and engineering, where understanding the total accumulation of a quantity is crucial.

Parabola

A parabola is a symmetrical, U-shaped curve on the graph, represented by a quadratic function. In the standard form, a parabola written as \( y = ax^2 + bx + c \) opens upward if the coefficient \( a \) is positive and downward if \( a \) is negative.

In our exercise, the curve is given by \( y = \frac{1}{4}(x^2 - 7) \), indicating the parabola opens upwards because of the positive coefficient \( \frac{1}{4} \). Parabolas are a common feature in problems involving calculus and coordinate geometry:

• Vertex: The peak or the lowest point of a parabola, which occurs at the axis of symmetry generated by the equation.
• Axis of Symmetry: A vertical line passing through the vertex, dividing the parabola into two equal halves.
• X-intercepts: Points where the parabola crosses the x-axis, found by setting \( y = 0 \) in the equation and solving for \( x \).

For this specific exercise, determining where the parabola intersects the x-axis involves solving \( \frac{1}{4}(x^2 - 7) = 0 \), which reveals the points at which it touches the x-axis, crucial for understanding the region of interest.

Area Between Curves

Finding the area between curves is a process of setting up an integral by determining the vertical distance between two curves over a specific section. The general idea is straightforward: subtract the bottom curve from the top curve and integrate over the given interval.

In our exercise, we aim to find the area between \( y = \frac{1}{4}(x^2 - 7) \) and \( y = 0 \) from \( x = 0 \) to \( x = 2 \).

• Identify the "top" and "bottom" curves; this means finding out which function lies above the other over the interval.
• Set up the integral of the difference: the integrand will be \( f(x) - g(x) \), where \( f(x) \) is the upper function and \( g(x) \) is the lower function.
• Compute the definite integral to find the total area.

Consider that the area where the top curve dips below the x-axis reverses the sign of the area computation, hence we take the absolute value for the final answer to ensure the area remains positive. This is critical as area cannot physically be negative.

Graph Sketching

Sketching graphs is a valuable skill in calculus as it visually represents the functions and assists with understanding relationships and intersections. Creating an accurate graph helps with setting up the problem and verifying your work.

When sketching a graph, consider the following:

• Determine the x- and y-intercepts of the function to understand where the curve crosses the axes.
• Identify the vertex or critical points for parabolas, aiding in understanding the shape and direction the graph takes.
• Plot several points to gain a sense of the curve's behavior between known intercepts and critical points.
• Check for symmetry which can simplify sketching, especially for parabolas and certain trigonometric functions.

In our exercise, sketching \( y = \frac{1}{4}(x^2 - 7) \) along with \( y = 0 \) helps distinguish the region where area calculations need to occur, clearly marking the boundaries from \( x=0 \) to \( x=2 \). This visual aid supports accurate integral setup and validation of computed areas.