Problem 15 - Thomas Calculus

Question

In Exercises \(15-22,\) graph the integrands and use areas to evaluate the integrals. $$ \int_{-2}^{4}\left(\frac{x}{2}+3\right) d x $$

Step-by-Step Solution

Answer

The integral evaluates to 21.

Step 1: Identify the Integrand

The integrand in the given exercise is \( f(x) = \frac{x}{2} + 3 \). This is a linear function.

Step 2: Graph the Integrand

To graph \( f(x) = \frac{x}{2} + 3 \), identify its y-intercept and slope. The y-intercept is when \( x = 0 \), so \( y = 3 \). The slope is \( \frac{1}{2} \), meaning for every 1 unit increase in \( x \), \( y \) increases by 0.5 units. Plot the y-intercept at (0,3) and another point by moving 2 units right (\( x = 2 \)) which gives \( y = 4 \). Draw a line through these points.

Step 3: Determine Limits of Integration

The given integral has limits from \( x = -2 \) to \( x = 4 \). On the graph, you need to consider the area under the line between these x-values.

Step 4: Calculate the Area Under the Graph

The shape formed under the line between \( x = -2 \) and \( x = 4 \) is a trapezoid. The formula for the area of a trapezoid is \( \frac{1}{2} \times (b_1 + b_2) \times h \), where \( h \) is the height (or the distance between \( x = -2 \) and \( x = 4 \)), and \( b_1 \) and \( b_2 \) are the lengths of the parallel sides (\( f(-2) = 2 \) and \( f(4) = 5 \)).

Step 5: Find the Function Values at the Limits

Calculate \( f(-2) = \frac{-2}{2} + 3 = 2 \) and \( f(4) = \frac{4}{2} + 3 = 5 \). These are the lengths of the parallel sides of the trapezoid (\( b_1 \) and \( b_2 \)).

Step 6: Apply the Area Formula

Plug the values into the trapezoid area formula:\[ \text{Area} = \frac{1}{2} \times (2 + 5) \times (4 - (-2)) = \frac{1}{2} \times 7 \times 6 = 21.\]

Step 7: Conclude the Solution

The value of the integral \( \int_{-2}^{4}\left(\frac{x}{2}+3\right) dx \) is 21. This is the area under the line from \( x = -2 \) to \( x = 4 \).

Key Concepts

Area Under the Curve

Understanding the concept of the "area under the curve" is crucial when dealing with definite integrals. It represents the total sum of all values a function takes over an interval, and it's the graphical representation of integration.
In this exercise, we work with a linear function given by the equation \( f(x) = \frac{x}{2} + 3 \). By calculating the area under this line from \( x = -2 \) to \( x = 4 \), we can evaluate the definite integral. This area forms a geometric shape known as a trapezoid.

To find the area of such a trapezoid, we use the formula:

• \( b_1 \) and \( b_2 \) are the function values at \( x = -2 \) and \( x = 4 \), respectively.
• The width or height \( h \) is the distance between these x-values: \( 4 - (-2) = 6 \).

The trapezoid's area is then calculated and represents the integral's value between these bounds. In essence, the definite integral isn't just a number; it's the total area caught between the curve and the x-axis over a specified interval.

Linear Functions

Linear functions have a constant rate of change, which is evident through their simple algebraic form \( y = mx + c \). Here, \( m \) is the slope, and \( c \) is the y-intercept.
In our integrand function \( f(x) = \frac{x}{2} + 3 \), the slope \( \frac{1}{2} \) describes how steep the line is, and the y-intercept 3 tells us where the line crosses the y-axis.

To graph a linear function, you first plot the y-intercept. Then, use the slope to find another point on the line:

• From the y-intercept, rise 0.5 units for every 1 unit you move right.
• Plot the next point accordingly and draw a line through them.

Linear functions are straightforward to graph and understand. They form straight lines that can be used to model consistent, uniform relationships in data or geometric problem setting.

Graphing

Graphing is a visual tool that helps us understand mathematical functions and their properties. It involves sketching the behavior of a function on a coordinate plane, showcasing how it changes with different x-values.
For the function \( f(x) = \frac{x}{2} + 3 \), begin by pinpointing critical elements:

• Locate the y-intercept: Here, it's at (0,3).
• Use the slope \( \frac{1}{2} \) to identify how the line progresses by rising 0.5 for each unit run to the right.

After marking enough points, you draw a straight line through them.

Graphing linear functions is especially helpful in illustrating the concept of "area under the curve" as seen in definite integrals. You can visually identify the area by the shape such as trapezoids, rectangles, or triangles formed beneath the line within specific intervals. This not only aids in evaluating integrals but also solidifies the understanding of geometric interpretation in calculus.