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Ch. 6 - Applications of Integration
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
Chapter 6, Problem 6.R.17

14–25. {Use of Tech} Areas of regions Determine the area of the given region.


The region bounded by y = ln x,y = 1, and x = 1

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Step 1: Understand the problem. The goal is to find the area of the region bounded by the curves y = ln(x), y = 1, and x = 1. This involves setting up an integral to calculate the area between these boundaries.
Step 2: Identify the limits of integration. The region is bounded by x = 1 and the vertical line where y = 1 intersects y = ln(x). To find this intersection, solve ln(x) = 1. This gives x = e (Euler's number). Thus, the limits of integration are from x = 1 to x = e.
Step 3: Set up the integral. The area is calculated by integrating the difference between the upper curve (y = 1) and the lower curve (y = ln(x)) over the interval [1, e]. The integral is: ∫x1e(1-ln(x))dx
Step 4: Break down the integral. The integral can be split into two parts: ∫x1e1dx and -∫x1eln(x)dx. Evaluate each part separately.
Step 5: Evaluate the integral. For the first part, ∫x1e1dx, the result is simply the length of the interval, e - 1. For the second part, -∫x1eln(x)dx, use integration by parts where u = ln(x) and dv = dx. Combine the results to find the total area.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Definite Integrals

Definite integrals are used to calculate the area under a curve between two points on the x-axis. In this context, the area of the region bounded by the curves can be found by integrating the function y = ln(x) from the lower limit x = 1 to the upper limit where y = 1 intersects the curve.
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Definition of the Definite Integral

Natural Logarithm Function

The natural logarithm function, denoted as ln(x), is the inverse of the exponential function e^x. It is defined for x > 0 and is crucial in this problem as it describes one of the boundaries of the region whose area we need to calculate.
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Derivative of the Natural Logarithmic Function

Area Between Curves

The area between curves is determined by finding the difference between the upper and lower functions over a specified interval. In this case, the area is calculated by integrating the difference between y = 1 and y = ln(x) from x = 1 to the point where ln(x) equals 1.
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Finding Area Between Curves on a Given Interval
Related Practice
Textbook Question

{Use of Tech} Decreasing velocity A projectile is fired upward, and its velocity (in m/s) is given by v(t) = 200 / √t+1, for t≥0.

a. Graph the velocity function, for t≥0.

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Surface area and volume Let f(x) = 1/3 x³ and let R be the region bounded by the graph of f and the x-axis on the interval [0, 2].


c. Find the volume of the solid generated when R is revolved about the x-axis.

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Textbook Question

43–55. Volumes of solids Choose the general slicing method, the disk/washer method, or the shell method to answer the following questions.


What is the volume of the solid whose base is the region in the first quadrant bounded by y = √x,y = 2-x, and the x-axis, and whose cross sections perpendicular to the base and parallel to the y-axis are semicircles?

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Textbook Question

Position, displacement, and distance A projectile is launched vertically from the ground at t=0, and its velocity in flight (in m/s) is given by v(t)=20−10t. Find the position, displacement, and distance traveled after t seconds, for 0≤t≤4.

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Textbook Question

82–84. Fluid Forces Suppose the following plates are placed on a vertical wall so that the top of the plate is 2 m below the surface of a pool that is filled with water. Compute the force on each plate.


A circular plate with a radius of 2 m

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Textbook Question

Surface area and volume Let f(x) = 1/3 x³ and let R be the region bounded by the graph of f and the x-axis on the interval [0, 2].


b. Find the volume of the solid generated when R is revolved about the y-axis.

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