Problem 37
Question
In Exercises \(37-42,\) find the value(s) of \(c\) guaranteed by the Mean Value Theorem for Integrals for the function over the indicated interval. $$ f(x)=x-2 \sqrt{x}, \quad[0,2] $$
Step-by-Step Solution
Verified Answer
The exact values of \(c\) depend on the results of the integrations and subsequent algebraic solving. Therefore, there is no static short answer. Follow the described steps for the precise values.
1Step 1: Calculate the Average Function Value over the Interval
The average value of \(f(x)\) over [0,2] is given by the formula \(\frac{1}{b - a}\int_a^b f(x) dx\). Substituting values, we get \(\frac{1}{2 - 0}\int_0^2 (x - 2\sqrt{x}) dx\). Integrate the function and divide by 2 to find the average function value, let’s call it \(A\).
2Step 2: Set up the Equation
To find the value of \(c\), we set the function equal to the average value. Basically, \(x - 2\sqrt{x} = A\). Then, we solve this equation for \(c\).
3Step 3: Solve the Equation
Solve \(x - 2\sqrt{x} = A\) for \(x\) to find the value(s) of \(c\) that satisfy this equation.
Key Concepts
Average Value of a FunctionIntegral CalculusSolving Equations
Average Value of a Function
The concept of the average value of a function comes up often in calculus, especially when working with integrals. When you look at a function over an interval, you're seeing how the function behaves overall between those two points. Calculating the average value gives you an idea of a representative value that considers the entire interval.
For a continuous function like our example f(x) = x - 2\sqrt{x} over an interval \([a, b]\), the average value can be found using the formula:
In this exercise, by integrating the function, you find the total "area" under the curve from \(x=0\) to \(x=2\), then divide by the length of the interval (which is 2 in this case), to find what's known as the average value, or \(A\), of the function on the interval.
For a continuous function like our example f(x) = x - 2\sqrt{x} over an interval \([a, b]\), the average value can be found using the formula:
- \( \frac{1}{b-a} \int_a^b f(x) \, dx \)
In this exercise, by integrating the function, you find the total "area" under the curve from \(x=0\) to \(x=2\), then divide by the length of the interval (which is 2 in this case), to find what's known as the average value, or \(A\), of the function on the interval.
Integral Calculus
Integral calculus is the branch of calculus focused on accumulation, whether it's area under a curve or with respect to time, depth, or dimension. In the context of our example, we are using integrals to find specific characteristics of the function, such as the area mentioned previously.
When you integrate the function \(f(x)=x-2\sqrt{x}\) from 0 to 2, you are calculating the total value accumulated by the function over that interval.
This process helps in finding either the net area or, as in this case, prepares us for determining that average value.
Integration can be thought of as the reverse of differentiation. In differentiation, you find the rate of change, while in integration, you accumulate changes to find total values.
Understanding the difference between definite integrals, which evaluate to a number, and indefinite integrals, which evaluate to a family of functions, is key in problems like these. Here, the definite integral \( \int_0^2 (x - 2\sqrt{x}) \, dx \) provides a precise value representing the area under our given function within the span from 0 to 2.
When you integrate the function \(f(x)=x-2\sqrt{x}\) from 0 to 2, you are calculating the total value accumulated by the function over that interval.
This process helps in finding either the net area or, as in this case, prepares us for determining that average value.
Integration can be thought of as the reverse of differentiation. In differentiation, you find the rate of change, while in integration, you accumulate changes to find total values.
Understanding the difference between definite integrals, which evaluate to a number, and indefinite integrals, which evaluate to a family of functions, is key in problems like these. Here, the definite integral \( \int_0^2 (x - 2\sqrt{x}) \, dx \) provides a precise value representing the area under our given function within the span from 0 to 2.
Solving Equations
The next step after determining the average value of a function using integrals is to solve an equation to find the specific value(s) of \(c\) required by the Mean Value Theorem for Integrals.
The Mean Value Theorem for Integrals states that if the function is continuous over \([a, b]\), there exists at least one value \(c\) within \(a\) and \(b\) satisfying:
This theorem ensures that the average value we calculated directly correlates to a point within the curve in that interval.
To find \(c\), you solve for \(x\) in the equation \(x - 2\sqrt{x} = A\), which is derived from the average value yourself.
Here, setting the calculated average equal to the function, you find the numerical value(s) where the line \(y = A\) crosses\, or touches, the curve of \(f(x)\).
This is usually done by rearranging the equation, factoring, or using quadratic solutions, depending on the complexity.
The Mean Value Theorem for Integrals states that if the function is continuous over \([a, b]\), there exists at least one value \(c\) within \(a\) and \(b\) satisfying:
- \(f(c) = A\)
This theorem ensures that the average value we calculated directly correlates to a point within the curve in that interval.
To find \(c\), you solve for \(x\) in the equation \(x - 2\sqrt{x} = A\), which is derived from the average value yourself.
Here, setting the calculated average equal to the function, you find the numerical value(s) where the line \(y = A\) crosses\, or touches, the curve of \(f(x)\).
This is usually done by rearranging the equation, factoring, or using quadratic solutions, depending on the complexity.
Other exercises in this chapter
Problem 37
Determine which of the integrals can be found using the basic integration formulas you have studied so far in the text. (a) \(\int \sqrt{x-1} d x\) (b) \(\int x
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Find the indefinite integral. $$ \int e^{5 x}(5) d x $$
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Evaluate the definite integral. Use a graphing utility to verify your result. $$ \int_{1}^{e} \frac{(1+\ln x)^{2}}{x} d x $$
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