Problem 7 - Calculus of a Single Variable

Question

Finding a Region In Exercises \(7-12\) , the integrand of the definite integral is a difference of two functions. Sketch the graph of each function and shade the region whose area is represented by the integral. $$ \int_{0}^{4}\left[(x+1)-\frac{x}{2}\right] d x $$

Step-by-Step Solution

Answer

The shaded area under the curve from \( x = 0 \) to \( x = 4 \) is 8 square units.

Step 1: Illustrate the Functions

The first function is \( y_1 = x + 1 \) which is a straight line with a slope of 1 and y-intercept of 1. The second function is \( y_2 = \frac{x}{2} \) which is a straight line with a slope of 0.5. Draw both functions on the same graph.

Step 2: Identify the Region Between the Functions

The region whose area is represented by the integral is the region between these two functions from \( x = 0 \) to \( x = 4 \). So, shade this region on the graph to visualize it better.

Step 3: Integrate to Find the Area

The area under the curve, between \( x = 0 \) and \( x = 4 \), for a function \( f(x) \) can be found by evaluating the definite integral \( \int_{0}^{4} f(x) dx \). Therefore, we have the integral \( \int_{0}^{4} ((x+1)-\frac{x}{2}) dx \). Break this into two integrals, find antiderivatives and evaluate at \( x = 4 \) and \( x = 0 \) and subtract: \(\int_{0}^{4} (x+1) dx - \int_{0}^{4} \frac{x}{2} dx \). Calculate the values of these integrals to get the area.

Key Concepts

Integrating Linear Functions

When we talk about integrating linear functions, we are referring to the process of finding the area under the graph of a function that can be represented by a straight line. A linear function has the general form of \( y = mx + b \), where \( m \) is the slope, and \( b \) is the y-intercept.

Integrating a linear function is a fundamental skill in calculus, as it allows us to calculate the total accumulated change over an interval. For example, in the given exercise, we integrate the linear functions \( y_1 = x + 1 \) and \( y_2 = \frac{x}{2} \). Using the power rule, the integral of \( y_1 \) with respect to \( x \) over the interval \( [0, 4] \) can be calculated as follows: \[ \int_{0}^{4} (x+1) dx = \frac{x^2}{2} + x \Big|_0^4 \] which simplifies to the evaluation of these expressions at \( x = 4 \) and subtracting the result when \( x = 0 \).

Area Under a Curve

The concept of the 'area under a curve' refers to the region enclosed by the graph of a function, the x-axis, and the vertical lines corresponding to the bounds of integration. To visualize this, consider the region that lies beneath the curve from one point to another along the x-axis.

The area can be positive or negative, depending on whether the curve lies above or below the x-axis over the interval. In the given exercise, we are interested in the area between two curves. By integrating the difference of the functions \( y_1 = x + 1 \) and \( y_2 = \frac{x}{2} \), we can determine this enclosed area.

Definite Integral

The definite integral is a fundamental concept used to calculate the exact area under a curve between two specific points. It is defined for a function \( f(x) \) over an interval \( [a, b] \) and is represented symbolically as \( \int_{a}^{b} f(x) dx \). This integral has a precise value, which is the net area, taking into account the above and below the x-axis distinctions.

For our exercise, the definite integral \( \int_{0}^{4}\big[(x+1)-\frac{x}{2}\big] dx \) will give us the exact area of the shaded region between the functions. After integrating, we evaluate the antiderivatives at the upper bound \( x = 4 \) and subtract the evaluation at the lower bound \( x = 0 \). This process yields the numerical value representing the area.

Graphical Representation of Functions

Understanding the graphical representation of functions is essential in visualizing concepts in calculus, particularly when calculating areas under curves or between functions. A graph allows us to see where a function lies in relation to the x-axis and how it interacts with other functions.

In the step-by-step solution provided, we use the graphs of two linear functions \( y_1 \) and \( y_2 \) to see the intersection points and the region of interest. The area of the shaded portion between these two functions from \( x = 0 \) to \( x = 4 \) is what we aim to calculate. By sketching these functions and shading the region, we establish a visual context which immensely aids in understanding the integration process and the computation of the definite integral.