Problem 23 - Calculus of a Single Variable

Question

Finding the Area of a Region In Exercises \(17-30,\) sketch the region bounded by the graphs of the equations and find the area of the region. $$ f(x)=\sqrt{x}+3, \quad g(x)=\frac{1}{2} x+3 $$

Step-by-Step Solution

Answer

The area of the region bounded by the graphs of the given equations is \(A=\frac{2}{3}(4) - (4) = \frac{8}{3} - 4 = -\frac{4}{3}\) square units. Since area cannot be negative, we take the absolute value, yielding \(\frac{4}{3}\) square units.

Step 1: Finding Points of Intersection

The Intersection points are where the given functions \(f(x)\) and \(g(x)\) meet. Let's equate the two functions and solve for \(x\): \(\sqrt{x}+3 = \frac{1}{2} x+3\). Solve this to find the \(x\) coordinates of the intersection points.

Step 2: Solving the Equation

First, we subtract 3 from both sides of the equation \(\sqrt{x}+3 = \frac{1}{2} x+3\), which gives us \(\sqrt{x} = \frac{1}{2}x\). To get rid of the square root, square both sides of the equation: \(x= (\frac{1}{2} x)^2\). From this, we can see that \(x=0\) or \(x=4\). So the intersection points are \(x=0\) and \(x=4\).

Step 3: Sketch the Graphs

Plot the functions \(f(x)=\sqrt{x}+3\) and \(g(x)=\frac{1}{2} x+3\) on the same graph. The area of interest is the region between these two functions from \(x=0\) to \(x=4\).

Step 4: Finding the Area

To find the area between these two functions from \(x=0\) to \(x=4\), subtract one function from the other and integrate from \(x=0\) to \(x=4\). Hence the area \(A\) can be given as: \[A =\int_{0}^{4}(f(x) - g(x))dx = \int_{0}^{4}(\sqrt{x}+3 - (\frac{1}{2}x+3))dx\]

Step 5: Simplify the Integration

Now we simplify and integrate the expression. That gives us: \[A=\int_{0}^{4}(\sqrt{x} - \frac{1}{2}x)dx\]

Step 6: Compute the Integral

Now compute the integral: \[A =\frac{2}{3}x^{3/2}|_{0}^{4} - \frac{1}{4}x^2|_{0}^{4} = \frac{2}{3}(4) - (4)\]

Key Concepts

Integration Techniques

In calculus, integration is a fundamental concept used to find areas, volumes, central points, and many useful things. Integration techniques refer to the various methods used to perform integrals, which are mathematical expressions that give the area under a curve. There are multiple techniques such as substitution, integration by parts, partial fraction decomposition, and trigonometric integration which help solve more complex integrals.

For simpler functions, like the ones in our exercise \( f(x)=\sqrt{x}+3 \) and \( g(x)=\frac{1}{2} x+3 \), the basic integration rules apply. For instance, the power rule which states that the integral of \( x^n \) is \( \frac{1}{n+1}x^{n+1} \) when \( n eq -1 \) is used. In our case, integrating the function \( \sqrt{x} - \frac{1}{2}x \) involves applying the power rule to each term individually after expressing \( \sqrt{x} \) as \( x^{1/2} \) to suit the rule's format.

Definite Integrals

A definite integral has start and end values, called the limits of integration, and it gives a number as an answer. It represents the total area under a curve between two points on the x-axis. The area can be positive, negative, or zero, depending on the function and the interval.

The format of a definite integral is written as \( \int_{a}^{b}f(x)dx \) where \( [a, b] \) is the interval and \( f(x) \) is the function. In the exercise, we are asked to calculate the area between two curves from \( x=0 \) to \( x=4 \) using a definite integral. A key point to remember when calculating the area between two curves is to subtract the lower function from the upper function before integrating, which is expressed as \( \int_{0}^{4}(f(x) - g(x))dx \) in our case.

Functions and Graphs

To understand the relationships between functions and their graphs, it's important to visualize how equations correspond to shapes on a coordinate plane. In our exercise, the functions provided are \( f(x) \) and \( g(x) \), which correspond to a square root function and a linear function, respectively.

Functions reveal their nature through their graphs. For instance, the graph of \( f(x)=\sqrt{x}+3 \) will start at (0,3) and curve upwards because the square root function generally forms a parabolic shape. On the other hand, \( g(x) \) is a straight line with a slope of 1/2 and a y-intercept of 3. Visualizing these graphs helps us identify the area that requires calculation and apply the correct integration technique. Sketching graphs can lead to a better comprehension of the relationship between different functions and the regions they bound when intersecting.

Points of Intersection

The points of intersection of two or more graphs are found where their equations hold the same values simultaneously for both the x and y coordinates. These points are critical for finding the boundaries of the area between curves.

To find points of intersection, we set the two equations equal to each other and solve for the variable, which in our example, involves finding the solution to \( \sqrt{x}+3 = \frac{1}{2} x+3 \). Once the common points are determined, which for our exercise are \( x=0 \) and \( x=4 \) by solving \( x= (\frac{1}{2} x)^2 \), we can use these as the limits of integration. Knowing the points of intersection simplifies the calculation of the area between two curves within those limits, ensuring accuracy and a clear understanding of the function's behavior within that specific region.