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Chapter 27: Problem 10

Find the area between the curve \(y=\ln x\) and the \(x\) -axis for \(1 \leq x \leq 10\). Get an exact answer. (Hint: Slice the area perpendicular to the \(y\) -axis so that the height of each slice is \(\Delta y\). Use this to arrive at an integral that you can evaluate exactly.)

Short Answer

Expert verified
The area between the curve \(y=\ln x\) and the x-axis for \(1 \leq x \leq 10\) is 9 square units.

Step by step solution

01

Understand the integral

The integral of a function f(x) from a to b on the interval [a, b] represents the signed area under the curve y=f(x) and between x=a and x=b. In this case, the function is \(y=\ln x\) and our interval is [1, 10]. We want to find the area under the curve and above the x-axis, so we are looking for the positive area.
02

Set up the integral

We're given the suggestion to slice the area perpendicular to the y-axis. This means we're going to integrate with respect to y. To find the area, we rewrite x in terms of y. Given \(y = \ln x\), we can write \(x = e^y\). So the width of each slice is \(e^y\) and the height is the change in y, so that the area of each slice is \(\Delta A = e^y* \Delta y\). We sum up these small areas from y=\(\ln 1=0\) to y = \(\ln 10\). Therefore our expression becomes \(\int_0^{\ln 10} e^y dy\).
03

Evaluate the integral

The definite integral of \(e^y\) with respect to y from 0 to the natural logarithm of 10 can be evaluated directly, as the antiderivative of \(e^y\) is \(e^y\). Thus, we can calculate it as \([e^y]_0^{\ln 10}\). Substituting the upper and lower limits into our antiderivative gives us \(e^{\ln 10} - e^0\), which simplifies to 10 - 1, due to the property that \(e^{\ln a} = a\) and \(e^0 = 1\).
04

Final result

After calculating the expression, the area under the curve is found to be 10 - 1, which gives us a final area of 9 square units.

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Key Concepts

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

Natural Logarithm Function
The natural logarithm function, denoted as \( \ln x \), is one of the fundamental mathematical functions. It's the inverse of the exponential function with base \( e \), where \( e \) is an irrational constant approximately equal to 2.718. This function is crucial in the field of calculus as it simplifies the process of integration and differentiation involving exponential growth or decay.

  • The natural logarithm of 1, \( \ln 1 \), equals 0 because \( e^0 = 1 \).
  • It increases slowly and continuously as \( x \) increases.
  • It's undefined for non-positive values of \( x \), meaning you cannot take the logarithm of zero or negative numbers.
This function is often used in solving calculus problems that involve growth, such as population models or interest calculations.
Definite Integral
The concept of a definite integral is central in calculus. It is used to determine the total accumulation of a quantity, such as area under a curve, over a specific interval. With a definite integral, we're not just looking at the general antiderivatives but finding the precise accumulation between two points on the x-axis.
  • The integral sign \( \int \) represents the operation of integration.
  • The limits of integration, such as \( a \) and \( b \) in \( \int_a^b f(x) dx \), specify the interval over which you accumulate the values.
  • For our problem, calculating \( \int_0^{\ln 10} e^y dy \) gives the exact area between the curve and the x-axis.
Notably, the definite integral gives us the net area between the curve and the x-axis, accounting for areas where the curve might dip below the axis as negative.
Area Under a Curve
Finding the area under a curve is one of the most fundamental applications of a definite integral in calculus. The task involves calculating the space between the curve of a function and the x-axis over a specific interval.

To find the area under the graph of \( y = \ln x \) between \( x = 1 \) and \( x = 10 \), we convert the function into its inverse. This helps when integrating perpendicular to the y-axis, transforming the task into evaluating \( \int_0^{\ln 10} e^y dy \).
  • This integral essentially slices the area into infinitesimally small strips parallel to the y-axis, calculating the contribution of each to the total area.
  • The final result gives an exact measurement of the area, expressed in square units.
  • Thus, even complex curves can be analyzed rigorously with a definite integral.
Exponential Function
The exponential function, particularly the natural exponential function \( e^x \), plays a vital role in calculus, especially in problems involving apid growth or decay. This function can describe phenomena such as radioactive decay or population growth.

In our problem, the curve \( y = \ln x \) was iterated through its inverse: the exponential function. Here's how it works:
  • Given \( y = \ln x \), rewriting gives \( x = e^y \) because \( e^{\ln x} = x \).
  • Utilizing this inverse property allows us to integrate with respect to each variable conveniently.
  • The integral of \( e^y \) with respect to y results in the simple function \( e^y \).
Due to the distinct properties of exponential functions, they are indispensable in various fields of physics, engineering, and mathematics.

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Most popular questions from this chapter

The region \(A\) in the first quadrant is bounded by \(y=2 x, y=-3 x+10\), and \(y=-\frac{1}{9}\left(x^{2}-6 x\right) .\) It has corners at \((0,0),(2,4)\), and \((3,1) .\) Express the area of \(A\) is the sum or difference of definite integrals. You need not evaluate.

A circus tent has cylindrical symmetry about its center pole. The height a distance of \(x\) feet from the center pole is given by \(h(x)=\frac{8}{1+\frac{x^{2}}{16}}\) feet. What is the volume enclosed by the tent of radius \(4 ?\)

Liquid is being stored in a large spherical tank of radius 2 meters. The tank is completely full and has been left standing for a long time. A mineral suspended in the liquid is setting. Its density at a depth of \(h\) meters from the top is given by \(5 h\) milligrams per cubic meter. Determine the number of milligrams of the mineral contained in the tank.

Traditionally, when a college football team seems certain to receive a bid to play in the post-season Orange Bowl, fans begin to throw oranges onto the field. Suppose that at one point during a game, the number of oranges per square yard between the goal line and the 30 -yard line is given by \(\rho(x)=\frac{30-x}{3}\) oranges per square yard, where \(x\) is the number of yards from the goal line. If the field is 160 feet ( \(160 / 3\) yards) wide, how many oranges lie between the goal line and the 30 -yard line?

Find the area bounded below by the \(x\) -axis, and laterally by \(y=\ln x\), and the line segment joining \((e, 1)\) to \((2 e, 0)\).
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