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Chapter 5: Problem 74

Show that for any number \(a>0\), $$ \int_{1}^{a} \frac{1}{x} d x=\ln a $$ This equation is often used as a definition of natural logarithms, defining \(\ln a\) as the area under the curve \(y=1 / x\) between 1 and \(a\). \([\) Hint \(: \ln 1=0 .]\)

Short Answer

Expert verified
The integral \( \int_{1}^{a} \frac{1}{x} \, dx = \ln a \) evaluates to \( \ln a \).

Step by step solution

01

Understand the Integral

The problem requires us to compute the definite integral \( \int_{1}^{a} \frac{1}{x} \, dx \). This represents the area under the curve \( y = \frac{1}{x} \) from \( x = 1 \) to \( x = a \). We aim to show that this area equals \( \ln a \).
02

Recall the Fundamental Theorem of Calculus

To solve the integral, we use the Fundamental Theorem of Calculus, which states that if \( F(x) \) is an antiderivative of \( f(x) \), then the definite integral of \( f(x) \) from \( a \) to \( b \) is \( F(b) - F(a) \). In this scenario, we need to determine an antiderivative of \( \frac{1}{x} \).
03

Determine the Antiderivative

We need to find a function \( F(x) \) such that \( F'(x) = \frac{1}{x} \). The derivative of \( \ln x \) is \( \frac{1}{x} \), so \( F(x) = \ln x \) is an antiderivative of \( \frac{1}{x} \).
04

Evaluate the Definite Integral

We apply the Fundamental Theorem of Calculus: \[\int_{1}^{a} \frac{1}{x} \, dx = [ \ln x ]_{1}^{a} = \ln a - \ln 1.\]Given that \( \ln 1 = 0 \), as per the hint provided, the integral simplifies to \( \ln a - 0 = \ln a \).
05

Conclusion

Thus, we have shown that the area under the curve \( y = \frac{1}{x} \) from \( x = 1 \) to \( x = a \) is \( \ln a \), as required.

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Key Concepts

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

Natural Logarithms
Natural logarithms are closely linked to the study of exponential growth and decay. They are logarithms to the base of Euler's number, approximately equal to 2.71828.
Natural logarithms are denoted as \( \ln \), and are fundamental in calculus, particularly in describing growth processes and solving integrals that involve exponential functions.
  • Natural logarithms transform multiplicative relationships into additive ones. This property simplifies complex multiplications into easier additions.
  • The notation, \( \ln a \), refers specifically to the natural logarithm of \( a \), where \( a \) is a positive real number.
  • Natural logarithms have the unique property that \( \ln 1 = 0 \). This is evident in the exercise, showing that the area under \( y = \frac{1}{x} \) from 1 to 1 is zero.
This integral provides an elegant geometric representation of \( \ln a \) as the area under \( y = \frac{1}{x} \) from 1 to \( a \). This integral shows how exponential processes can be understood through areas under curves.
Fundamental Theorem of Calculus
The Fundamental Theorem of Calculus is a critical component in linking the concepts of differentiation and integration.
This theorem has two parts, but in this context, we focus on its application to evaluate definite integrals.
  • The theorem enables us to evaluate a definite integral \( \int_{a}^{b} f(x) \, dx \) by using the antiderivative \( F(x) \) of \( f(x) \). Specifically, it states \( F(b) - F(a) \).
  • With the function \( f(x) = \frac{1}{x} \), its antiderivative is \( F(x) = \ln x \) since the derivative of \( \ln x \) is \( \frac{1}{x} \).
  • By applying this theorem, we find that the integral from 1 to \( a \) results in \( \ln a - \ln 1 \), simplifying to \( \ln a \) as \( \ln 1 = 0 \).
This theorem simplifies evaluating definite integrals significantly, transforming complex area calculations into straightforward evaluations of an antiderivative at specified bounds.
Antiderivative
An antiderivative, or indefinite integral, provides a way to "reverse" differentiation.
It is a function \( F(x) \) whose derivative \( F'(x) \) returns the original function \( f(x) \).
  • Finding the antiderivative is essential in solving integrals. It allows us to determine the accumulated area under a curve.
  • For \( f(x) = \frac{1}{x} \), the antiderivative \( \ln x \) captures this idea because \( \frac{d}{dx} (\ln x) = \frac{1}{x} \).
  • This concept is highlighted when taking the integral of \( f(x) \) over a range to evaluate the total area. It shows the transition from instantaneous rate \( \frac{1}{x} \) to a total area \( \ln x \).
Antiderivatives are fundamental in calculus as they bridge the gap between the derivative and the integral, allowing us to tackle a vast variety of problems in a systematic way.

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Most popular questions from this chapter

BUSINESS: Revenue An aircraft company estimates its marginal revenue function for helicopters to be \(M R(x)=(x+40) \sqrt{x^{2}+80 x}\) thousand dollars, where \(x\) is the number of helicopters sold. Find the total revenue from the sale of the first 10 helicopters.

Suppose that a company found its sales rate (in sales per day) if it did advertise, and also its (lower) sales rate if it did not advertise. If you integrated "upper minus lower" over a month, describe the meaning of the number that you would find.

Find the average value of each function over the given interval. $$ f(x)=e^{x / 2} \text { on }[0,2] $$

67-68. BUSINESS: Cumulative Profit A company's marginal revenue function is \(M R(x)=700 x^{-1}\) and its marginal cost function is \(M C(x)=500 x^{-1}\) (both in thousands of dollars), where \(x\) is the number of units \((x>1)\). Find the total profit from $$ x=100 \text { to } x=200 $$

The substitution method can be used to find integrals that do not fit our formulas. For example, observe how we find the following integral using the substitution \(u=x+4\) which implies that \(x=u-4\) and so \(d x=d u\). $$ \begin{aligned} \int(x-2)(x+4)^{8} d x &=\int(u-4-2) u^{8} d u \\ &=\int(u-6) u^{8} d u \\ &=\int\left(u^{9}-6 u^{8}\right) d u \\ &=\frac{1}{10} u^{10}-\frac{2}{3} u^{9}+C \\ &=\frac{1}{10}(x+4)^{10}-\frac{2}{3}(x+4)^{9}+C \end{aligned} $$ It is often best to choose \(u\) to be the quantity that is raised to a power. The following integrals may be found in this way (as well as by the methods of Section 6.1). $$ \int(x-1) \sqrt{x+2} d x $$
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