Chapter 2: Problem 24
Find the average value of \(f(x)=\left\\{\begin{array}{cc}x+4, & -4 \leq x
\leq-1 \\ -x+2, & -1
Short Answer
Expert verified
Answer: The average value of the function on the interval \([-4, 2]\) is 1.5.
Step by step solution
01
Divide the interval
Since the function is defined piecewise as \(x + 4\) in the interval \([-4, -1]\) and as \(-x + 2\) in the interval \((-1, 2]\), we should calculate the area under the curve for each part of the function.
02
Calculate the area under the curve for \(f(x) = x + 4\) in \([-4, -1]\)
To find the area under the curve for \(f(x) = x + 4\) on the interval \([-4, -1]\), we will calculate the area of the trapezoid formed by the graph of the function within this interval.
The vertices of the trapezoid are located at \((-4,0)\), \((-4, -4 + 4 = 0)\), \((-1, -1 + 4 = 3)\), and \((-1, 0)\). The height of the trapezoid is the length of the interval \(h = |-1 - (-4)| = 3\), and the bases are \(b_1 = f(-4) = 0\) and \(b_2 = f(-1) = 3\). Using the formula for the area of a trapezoid \(A = \frac{1}{2}(b_1 + b_2)h\), we get:
$$A_1 = \frac{1}{2}(0 + 3)(3) = \frac{1}{2}(3)(3) = 4.5$$
So the area under the curve in the interval \([-4, -1]\) is \(4.5\) square units.
03
Calculate the area under the curve for \(f(x) = -x + 2\) in \((-1, 2]\)
To find the area under the curve for \(f(x) = -x + 2\) on the interval \((-1, 2]\), we will calculate the area of the trapezoid formed by the graph of the function within this interval.
The vertices of the trapezoid are located at \((-1,0)\), \((-1, -(-1) + 2 = 3)\), \((2, -(2) + 2 = 0)\), and \((2, 0)\). The height of the trapezoid is the length of the interval \(h = |2 - (-1)| = 3\), and the bases are \(b_1 = f(-1) = 3\) and \(b_2 = f(2) = 0\). Using the formula for the area of a trapezoid \(A = \frac{1}{2}(b_1 + b_2)h\), we get:
$$A_2 = \frac{1}{2}(3 + 0)(3) = \frac{1}{2}(3)(3) = 4.5$$
So the area under the curve in the interval \((-1, 2]\) is \(4.5\) square units.
04
Add the areas and divide by the total interval length
To find the total area under the curve on the interval \([-4, 2]\), we add the two areas calculated in Steps 2 and 3:
$$A_{total} = A_1 + A_2 = 4.5 + 4.5 = 9$$
Finally, divide the total area by the length of the interval \((|2 - (-4)|)\):
$$\text{Average value}=\frac{A_{total}}{|(-4)-2|} = \frac{9}{6} = 1.5$$
So the average value of \(f(x)\) on the interval \([-4, 2]\) is \(1.5\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Piecewise Functions
Piecewise functions are unique in the sense that different parts of their domains have different rules or 'pieces' for determining the function's value. The function presented in the original exercise,
\(f(x)=\left\{\begin{array}{cc} x+4, & -4 \leq x \leq -1 \ -x+2, & -1
is a classic example of a piecewise function with two distinct pieces. To fully understand and work with such functions, it's critical to consider the specific interval that each piece applies to. In many real-world scenarios, such as calculating income tax or determining shipping charges, piecewise functions are employed to represent different rates or fees across various ranges. When visualizing piecewise functions, you'll typically see a graph that may have sharp turns or breaks, corresponding to the points where the function 'switches' from one rule to another.
\(f(x)=\left\{\begin{array}{cc} x+4, & -4 \leq x \leq -1 \ -x+2, & -1
is a classic example of a piecewise function with two distinct pieces. To fully understand and work with such functions, it's critical to consider the specific interval that each piece applies to. In many real-world scenarios, such as calculating income tax or determining shipping charges, piecewise functions are employed to represent different rates or fees across various ranges. When visualizing piecewise functions, you'll typically see a graph that may have sharp turns or breaks, corresponding to the points where the function 'switches' from one rule to another.
Area Under the Curve
The concept of 'area under the curve' is central to understanding integral calculus, but it can begin with the simple geometric idea of finding the area under a graph of a function over a certain interval. In the context of the exercise, calculating the area under the curve for each piece of the piecewise function gives us a geometric representation of the function's behavior over the interval
\([-4, 2]\).
For the visual thinkers, imagine coloring in the region directly under the function's graph, between the x-axis and the function's line. This region represents the total 'effect' or 'quantity' accumulated by the function over that interval. In many practical applications, such as physics and economics, the area under the curve can signify anything from distance traveled over time to total revenue over a price range. The combined area of these pieces helps us determine the average value without the use of integral calculus, making it a valuable approach for students first learning about these concepts.
\([-4, 2]\).
For the visual thinkers, imagine coloring in the region directly under the function's graph, between the x-axis and the function's line. This region represents the total 'effect' or 'quantity' accumulated by the function over that interval. In many practical applications, such as physics and economics, the area under the curve can signify anything from distance traveled over time to total revenue over a price range. The combined area of these pieces helps us determine the average value without the use of integral calculus, making it a valuable approach for students first learning about these concepts.
Trapezoidal Rule
The trapezoidal rule is a numerical method to approximate the area under a curve, which is a key step in our exercise. When applying the trapezoidal rule, we divide the area into trapezoids rather than rectangles—as is done in the more basic Riemann sums method—since trapezoids can better approximate the area under a curve that is not perfectly straight.
In the given problem, we used the trapezoidal rule to calculate the areas under two pieces of the piecewise function. To do this, we located the vertices of the trapezoids formed by the graphs of the function's pieces and the x-axis for each interval. We then applied the formula for the area of a trapezoid
\(A = \frac{1}{2}(b_1 + b_2)h\),
where \(b_1\) and \(b_2\) are the lengths of the parallel sides (or bases) of the trapezoid and \(h\) is the height. The use of trapezoids in this way allowed us to obtain a more accurate approximation of the area under each piece of the function compared to using rectangles, especially when dealing with linear functions as pieces of the piecewise function.
In the given problem, we used the trapezoidal rule to calculate the areas under two pieces of the piecewise function. To do this, we located the vertices of the trapezoids formed by the graphs of the function's pieces and the x-axis for each interval. We then applied the formula for the area of a trapezoid
\(A = \frac{1}{2}(b_1 + b_2)h\),
where \(b_1\) and \(b_2\) are the lengths of the parallel sides (or bases) of the trapezoid and \(h\) is the height. The use of trapezoids in this way allowed us to obtain a more accurate approximation of the area under each piece of the function compared to using rectangles, especially when dealing with linear functions as pieces of the piecewise function.