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

Short Answer

Expert verified
The area bounded between the x-axis, \(y =\ln x\), and the line from \((e, 1)\) to \((2e, 0)\) can be found by calculating the definite integral \(A = \int_{e}^{2e} (\ln x - ( -x + 1 + e))dx\).

Step by step solution

01

Calculate the equation of the line

Using the two points given, \((e, 1)\) and \((2e, 0)\), the slope \(m\) of the line can be calculated as: \(\frac{0 - 1}{2e - e} = -1\). Using the y-intercept formula \(b = y - mx\), with the point \((e, 1)\) we get \(b = 1 - (-1 \cdot e) = 1 + e\). Therefore, the equation of the line is \(y = -x + 1 + e\).
02

Find the intersection points

The area is bounded by the intersection points of the line and the function \(y=\ln x\). To find these intersection points, we can set the equations equal to each other: \(\ln x = -x + 1 + e\). By looking at the graph or through calculation, we can determine that the two intersection points are \(x=e\) and \(x=2e\).
03

Calculate the area

The area between two functions from a to b can be calculated as \(\int_{a}^{b}(f(x) - g(x))dx\), where \(f(x)\) and \(g(x)\) are the equations of the two functions. In our case, \(f(x) = \ln x\) and \(g(x) = -x + 1 + e\), and our boundaries are \(x=e\) and \(x=2e\). Thus, the area \(A\) is given by \(A = \int_{e}^{2e} (\ln x - ( -x + 1 + e))dx\). Calculating this integral yields the result.

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Key Concepts

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

Integrals in Calculus
Integrals in calculus are powerful tools that help us find areas under curves, among other things. They allow us to calculate the accumulated total of quantities, which can represent areas, volumes, and more. When dealing with integrals, it's important to understand that they work by summing up infinitely small pieces. This concept might sound complicated, but think of it as cutting a shape into tiny strips, and then adding those strip areas together.
Integrals come in two main types:
  • Indefinite integrals provide a family of functions and include a constant of integration.
  • Definite integrals, which are used when we're calculating specific values, like the area between two points.
The process of finding these integrals is known as integration, and it is the reverse operation of differentiation. By understanding the properties and methods of integration, we can solve complex problems related to the area bounded by curves in no time.
Equation of a Line
The equation of a line is a fundamental concept in algebra and is used frequently in calculus. It's a way of representing a straight line using a simple algebraic formula. Typically, a line in the plane can be described by the equation: \[ y = mx + b \]where:
  • \( m \) is the slope of the line, representing its steepness.
  • \( b \) is the y-intercept, which is the point where the line crosses the y-axis.

To find the equation of a line between two points, say \((x_1, y_1)\) and \((x_2, y_2)\), we first calculate the slope using \( m = rac{y_2 - y_1}{x_2 - x_1} \). Then, we plug a point into the line equation to solve for \( b \). These tools allow us to describe any line in a straightforward manner, making it easier to find where this line intersects with other curves.
Intersection Points
Intersection points are where two curves meet on a graph, and they're pivotal when calculating areas bounded by curves. By setting the two functions equal, we can solve for the \( x \)-coordinates where their graphs intersect. This step is crucial because these points will define the limits of integration.
Here's how you generally find intersection points:
  • Set the equations equal: \( f(x) = g(x) \).
  • Solve the equation to find the possible \( x \)-values.

Sometimes, finding these solutions requires algebraic manipulation or using numerical methods. However, once the intersection points are known, they can be used to set up the bounds for definite integrals to calculate the area you're interested in.
Definite Integrals
Definite integrals are a specific kind of integral used to calculate exact quantities, like the area under a curve between two specified points. Unlike indefinite integrals, definite integrals give a numerical value. The notation for a definite integral from \( a \) to \( b \) is:\[ \int_{a}^{b} f(x) \, dx \]Here’s what each part means:
  • \( a \) and \( b \) are the limits of integration, showing the interval over which you’re integrating.
  • The function \( f(x) \) is the curve whose area you're calculating.
  • \, dx represents a small change in \( x \).
When you evaluate a definite integral, you apply the Fundamental Theorem of Calculus. This involves finding the antiderivative of \( f(x) \), then calculating it at \( b \) and \( a \), and subtracting the two. This process gives you the exact area under \( f(x) \) between \( a \) and \( b \), which is very useful in a wide range of scientific and engineering applications.

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Most popular questions from this chapter

A substance has been put in a centrifuge. We now have a cylindrical sample (radius 3 centimeters, height 4 centimeters) in which density varies with \(x\), the distance (in centimeters) from the central axis. If the density is given by \(\rho(x) \mathrm{mg} / \mathrm{cm}^{3}\), write an integral that gives the total mass of the substance.

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.

Evaluate \(\int_{0}^{1} \arctan x d x\) by interpreting it as an area and slicing horizontally.

Let \(W(t)\) be the amount of water in a pool at time \(t, t\) measured in hours and \(W\) measured in gallons. \(t=0\) corresponds to noon. Water is flowing in and out of the pool at a rate given by \(\frac{d W}{d t}=30 \cos \left(\frac{\pi}{2} t\right)\). During what time interval between noon and \(5: 00\) P.M. \((0 \leq t \leq 5)\) is water flowing out of the pool at a rate of 15 gallons an hour or more? How much water actually has left the pool in this time interval?

A coastal town is in the shape of a 7 -mile by 2 -mile rectangle, with one of the 7 -mile sides along the coast. In this town people want to live near the beach and the population density at a distance \(x\) from the coast is given by \(\delta(x)=4000-2000 x\) people per square mile. (a) Write a general Riemann sum that approximates the total population of the town. (b) Use your answer to part (a) to write a definite integral that represents the total population of the town and evaluate the integral.

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