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Chapter 7: Problem 53

Find the average value of the function over the given interval. $$f(x)=1 / x ;[1, e]$$

Short Answer

Expert verified
The average value over the interval \([1, e]\) is \( \frac{1}{e-1} \).

Step by step solution

01

Understand the Formula for Average Value

The average value of a function \( f(x) \) over the interval \([a, b]\) is given by the formula:\[\text{Average Value} = \frac{1}{b-a} \int_{a}^{b} f(x) \, dx\]Here, \( a = 1 \) and \( b = e \). Our function is \( f(x) = \frac{1}{x} \).
02

Set Up the Integral

Begin by setting up the definite integral of \( f(x) \) over the interval from 1 to \( e \):\[\int_{1}^{e} \frac{1}{x} \, dx\]
03

Evaluate the Integral

The integral of \( \frac{1}{x} \) is the natural logarithm \( \ln |x| \). Evaluate the integral from 1 to \( e \):\[\left. \ln |x| \right|_1^e = \ln e - \ln 1\]Since \( \ln e = 1 \) and \( \ln 1 = 0 \), the value of the integral is \( 1 - 0 = 1 \).
04

Calculate the Average Value

Substitute the value of the integral from Step 3 into the average value formula:\[\text{Average Value} = \frac{1}{e-1} \times 1\]This simplifies to:\[\frac{1}{e-1}\]
05

Present the Final Answer

Hence, the average value of the function \( f(x) = \frac{1}{x} \) over the interval \([1, e]\) is \( \frac{1}{e-1} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Definite Integral
The definite integral is a fundamental concept in calculus. It is used to calculate the accumulation of quantities, such as area under curves. The notation for a definite integral is \( \int_{a}^{b} f(x) \, dx \), where \( a \) and \( b \) are the lower and upper bounds, respectively.

Here's what a definite integral does:
  • Calculates the net area under the curve \( f(x) \) from \( x = a \) to \( x = b \).
  • It gives us a specific, numerical value representing that area.
  • The integration process accumulates values of \( f(x) \) over the interval \([a, b]\).
For example, in the exercise, we evaluated \( \int_{1}^{e} \frac{1}{x} \, dx \), which helps in finding the area under the curve \( f(x) = \frac{1}{x} \) from \( x = 1 \) to \( x = e \). This step is essential because it provides the integral needed to calculate the average value of the function.
Average Value of a Function
The average value of a function over an interval gives us the "typical" value that the function takes in that interval. We compute it using the formula:\[\text{Average Value} = \frac{1}{b-a} \int_{a}^{b} f(x) \, dx\]This represents the mean height of the function over the interval \([a, b]\).

Why do we use this method?
  • It helps in understanding the overall behavior of the function across the interval.
  • It provides insights similar to statistical averages but in a continuous setting.
In our problem, the interval is \([1, e]\), and the function is \( f(x) = \frac{1}{x} \). After evaluating the integral, plugging the result into the average value formula gives us \( \frac{1}{e-1} \). This result reflects the average or typical value of the function \( \frac{1}{x} \) as we move from 1 to \( e \).
Natural Logarithm
The natural logarithm, commonly denoted as \( \ln \), is a logarithm to the base \( e \), where \( e \approx 2.71828 \). It is the inverse operation of the exponential function with base \( e \). The logarithm function is integral to solving many calculus problems.

Some properties of natural logarithms include:
  • \( \ln(e) = 1 \) because \( e^1 = e \).
  • \( \ln(1) = 0 \) because \( e^0 = 1 \).
  • Logarithms convert multiplication and division into addition and subtraction.
In our exercise, the natural logarithm appears during the evaluation of \( \int_{1}^{e} \frac{1}{x} \, dx \). The integral of \( \frac{1}{x} \) is \( \ln|x| \). When evaluated from 1 to \( e \), it gives us \( \ln(e) - \ln(1) = 1 - 0 = 1 \). Understanding how logarithms work is crucial for correctly calculating definite integrals involving functions like \( \frac{1}{x} \).

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Most popular questions from this chapter

Find the area under the curve \(y=3 \cos 2 x\) over the interval \([0, \pi / 8]\)

Use a CAS to find the exact value of the integral $$\int_{-3}^{1} \sqrt{3-2 x-x^{2}} d x$$ and then confirm the result by hand calculation. [Hint: Complete the square. \(]\)

(a) Give a geometric argument to show that $$ \frac{1}{x+1}<\int_{x}^{x+1} \frac{1}{t} d t<\frac{1}{x}, \quad x>0$$ (b) Use the result in part (a) to prove that $$ \frac{1}{x+1}<\ln \left(1+\frac{1}{x}\right)<\frac{1}{x}, \quad x>0$$ (c) Use the result in part (b) to prove that $$ e^{\frac{1}{\ln 4}}<\left(1+\frac{1}{x}\right)^{x}0$$ and hence that $$\lim _{x \rightarrow+\infty}\left(1+\frac{1}{x}\right)^{x}=e$$ (d) Use the inequality in part (c) to prove that $$\left(1+\frac{1}{x}\right)^{x}0$$

Evaluate the definite integral two ways: first by a \(u\) -substitution in the definite integral and then by a \(u\) -substitution in the corresponding indefinite integral. $$\int_{0}^{\ln 5} e^{x}\left(3-4 e^{x}\right) d x$$

Given that \(n\) is a positive integer, show that $$ \int_{0}^{\pi / 2} \sin ^{n} x d x=\int_{0}^{\pi / 2} \cos ^{n} x d x $$ by using a trigonometric identity and making a substitution. Do not attempt to evaluate the integrals.
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