Problem 31
Question
Let \(f(x)=2 x, 0 \leq x \leq 2\). Use a geometric argument to find the average value of \(f\) over the interval \([0,2]\), and find \(x\) such that \(f(x)\) is equal to this average value.
Step-by-Step Solution
Verified Answer
The average value of the function over [0,2] is 2, and for this, x is 1.
1Step 1: Understand the Function and the Interval
The function is given as \(f(x) = 2x\) for the interval \([0, 2]\). Our goal is to find the average value of this function over the interval and then the \(x\) such that \(f(x)\) is equal to this average value.
2Step 2: Visualize the Geometric Interpretation
In a geometric sense, the average value of a function over an interval can be interpreted as the height of a rectangle that has the same area as the area under the curve on that interval.
3Step 3: Compute the Area Under the Curve
The area under the function \(f(x) = 2x\) from \(x = 0\) to \(x = 2\) forms a triangle. The base of this triangle is 2, and the height is \(f(2) = 4\). Hence, the area \(A\) of the triangle is calculated as follows:\[ A = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 2 \times 4 = 4. \]
4Step 4: Calculate the Average Value of the Function
The average value of \(f(x)\) over the interval \([0,2]\) is the total area under the curve (which we've found as 4) divided by the length of the interval (which is 2):\[ f_{avg} = \frac{4}{2} = 2. \]
5Step 5: Find the x-Value Corresponding to the Average Value
We set \(f(x)\) equal to the average value to solve for \(x\):\[ f(x) = 2x = 2. \]Solving this equation for \(x\) gives:\[ x = 1. \]
Key Concepts
Average Value of a FunctionGeometric InterpretationArea Under a Curve
Average Value of a Function
In calculus, the concept of the average value of a function over an interval is an important one. It's similar to finding the average of a set of numbers, but instead, we apply it to functions. To understand this, imagine you are asked to find the average of several exam scores: you sum up all the scores and divide by the number of exams.
When finding the average value of a function, we follow a similar process, but we're calculating the average over a continuous interval rather than a set of discrete points. The average value of a function, denoted as \( f_{avg} \), over an interval \([a, b]\), is given by the formula:
\[ f_{avg} = \frac{1}{b-a} \int_a^b f(x) \, dx \]
In this equation:
When finding the average value of a function, we follow a similar process, but we're calculating the average over a continuous interval rather than a set of discrete points. The average value of a function, denoted as \( f_{avg} \), over an interval \([a, b]\), is given by the formula:
\[ f_{avg} = \frac{1}{b-a} \int_a^b f(x) \, dx \]
In this equation:
- \(b-a\) represents the length of the interval.
- \(\int_a^b f(x) \, dx\) is the integral of the function from \(a\) to \(b\), which gives us the total area under the curve.
Geometric Interpretation
In calculus, a geometric interpretation helps us visualize complex algebraic ideas using shapes and areas. For the average value of a function, this visualization becomes particularly useful. Imagine the graph of a function \(f(x)\) over an interval \([a, b]\).
The area under this curve represents the integral of \(f(x)\) from \(a\) to \(b\). Geometrically, the average value of the function over this interval corresponds to the height of a rectangle that has the same width as the interval and the same total area as the area under the curve.
For example, if the area under \(f(x)\) from \(x=0\) to \(x=2\) is 4, then to find the average value rectangle, we ensure that the rectangle's width is 2 (from 0 to 2) and its height is the average value \(f_{avg}\).
This method allows us to "average out" the peaks and valleys in the function graph, creating a simpler rectangle that helps in understanding the function's average performance across that interval.
The area under this curve represents the integral of \(f(x)\) from \(a\) to \(b\). Geometrically, the average value of the function over this interval corresponds to the height of a rectangle that has the same width as the interval and the same total area as the area under the curve.
For example, if the area under \(f(x)\) from \(x=0\) to \(x=2\) is 4, then to find the average value rectangle, we ensure that the rectangle's width is 2 (from 0 to 2) and its height is the average value \(f_{avg}\).
This method allows us to "average out" the peaks and valleys in the function graph, creating a simpler rectangle that helps in understanding the function's average performance across that interval.
Area Under a Curve
The area under a curve is a crucial concept in calculus, especially when we're dealing with integrals. When you hear "area under the curve," think of it as the space between the curve of \(f(x)\) and the x-axis, over a specified interval.
This area can represent various physical and mathematical concepts. For instance:
This area can represent various physical and mathematical concepts. For instance:
- In physics, it might represent the total distance covered by an object if the curve represents a velocity function over time.
- In economics, it could symbolize total revenue if the curve indicates a revenue function based on sales volume.
Other exercises in this chapter
Problem 30
Use the algebraic rules for sums to evaluate each sum. Recall that $$\sum_{k=1}^{n} k=\frac{n(n+1)}{2}$$ and $$\sum_{k=1}^{n} k^{2}=\frac{n(n+1)(2 n+1)}{6}$$ $$
View solution Problem 31
Use Leibniz's rule to find \(\frac{d y}{d x}\). $$ y=\int_{x^{2}}^{1} \sec t d t,-1
View solution Problem 32
Use Leibniz's rule to find \(\frac{d y}{d x}\). $$ y=\int_{2+x^{2}}^{2} \cot t d t $$
View solution Problem 32
A particle moves along the \(x\) -axis with velocity $$ v(t)=-(t-3)^{2}+5 $$ for \(0 \leq t \leq 6\). (a) Graph \(v(t)\) as a function of \(t\) for \(0 \leq t \
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