Problem 16 - University Calculus: Early Transcendentals

Question

Graph the integrands and use known area formulas to evaluate the integrals. $$\int_{1 / 2}^{3 / 2}(-2 x+4) d x$$

Step-by-Step Solution

Answer

The integral evaluates to 2.25.

Step 1: Understand the Problem

We need to evaluate the integral \( \int_{1 / 2}^{3 / 2}(-2x+4) \, dx \). To solve this, we will graph the function \( f(x) = -2x+4 \) over the interval \( [1/2, 3/2] \) and compute the area under the curve using geometric formulas.

Step 2: Sketch the Function

The function \( f(x) = -2x + 4 \) is a straight line. It has a slope of \(-2\) and a y-intercept of \(4\). To graph it, note that when \( x = 0 \), \( f(x) = 4 \), and when \( x = 2 \), \( f(x) = 0 \). Plot these points and draw a line through them.

Step 3: Identify the Relevant Portion of the Graph

We are interested in the section of the graph between \( x = 1/2 \) and \( x = 3/2 \). At \( x = 1/2 \), \( f(x) = -1 + 4 = 3.5 \), and at \( x = 3/2 \), \( f(x) = -3 + 4 = 1 \). Mark these points on the line.

Step 4: Determine the Area Using Geometry

The segment of the line \( f(x) = -2x + 4 \) between \( x = 1/2 \) and \( x = 3/2 \) forms a trapezoid with the x-axis. The trapezoid has bases \( f(1/2) = 3 \) and \( f(3/2) = 1 \) with height \( 3/2 - 1/2 = 1 \). Use the formula for the area of a trapezoid: \( \frac{1}{2} \times (b_1 + b_2) \times h \).

Step 5: Calculate the Area of the Trapezoid

Plug into the trapezoid area formula: \[ \text{Area} = \frac{1}{2} \times (3.5 + 1) \times 1 = \frac{1}{2} \times 4.5 \times 1 = 2.25. \] Thus, the value of the integral is 2.25.

Key Concepts

Trapezoidal Rule

When it comes to estimating the area under a curve, the Trapezoidal Rule is a handy method to have in your toolkit. This rule helps calculate an integral by dividing the area into trapezoids rather than the traditional rectangles. The area of a trapezoid is calculated using the formula: - \[ \frac{1}{2} \times (b_1 + b_2) \times h \] - Where \( b_1 \) and \( b_2 \) are the lengths of the parallel sides of the trapezoid, and \( h \) is the height. In our exercise, we used this rule with a linear function \( f(x) = -2x + 4 \) between the limits of \( \frac{1}{2} \) and \( \frac{3}{2} \). This function, when plotted, created a clear trapezoid shape, facilitating an easier computation of the area, thus evaluating the definite integral accurately.

Graphing Functions

Graphing functions is an essential skill in calculus and mathematics in general. It provides a visual representation of relationships defined by formulas. For linear functions like \( f(x) = -2x + 4 \), graphing reveals its linearity and helps identify important characteristics:

• Slope: The coefficient of \( x \) in the function, here \(-2\), represents the slope. It tells us that for every unit increase in \( x \), the function decreases by 2 units.
• Y-intercept: This is the constant term in the function, 4 in our case, indicating where the line crosses the y-axis.

When graphing this function, it's helpful to find values of \( x \) that result in convenient outputs for \( f(x) \) and plot those points to draw the straight line.

Linear Functions

A linear function is characterized by a straight-line graph. These functions are expressed in the form \( f(x) = mx + b \), where \( m \) is the slope, and \( b \) is the y-intercept. They are straightforward but powerful since many real-world situations can be simplified into linear models. Some key features of linear functions include:

• Constant Slope: The rate of change is uniform across any two points on the line.
• No Curvature: Unlike quadratic or cubic functions, linear functions have no bends or twists.
• Solving and Factorization: Linear equations can typically be rearranged and solved without complicated algebra.

Understanding these properties is crucial when calculating areas under their graphs or assessing their behavior over specific intervals.

Area Under a Curve

Calculating the area under a curve is a fundamental task in calculus that usually involves integrating a function over a specific interval. The concept is based on summing infinitely small slices of area under a curve defined by a function. For simple shapes, like trapezoids formed between a curve and the x-axis (as seen in our exercise), geometric formulas can be used instead of lengthy integrations. Using graphical methods such as the trapezoidal rule to find the area helps visualize the integral as more than just numbers and equations. It shows the physical space covered under the curve, giving a better intuitive understanding of what an integral represents. This is especially more evident when dealing with linear functions, where the area between the graph and the x-axis forms a simple trapezoid. Calculating these manually reinforces the conceptual bridge between algebraic operations and geometric space.