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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
All textbooksBriggs 3rd EditionCh. 6 - Applications of IntegrationProblem 6.2.61
Chapter 6, Problem 6.2.61

Find the area of the region described in the following exercises.


The region in the first quadrant bounded by y=5/2−1/x and y=x

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First, identify the curves that bound the region: the line \(y = x\) and the curve \(y = \frac{5}{2} - \frac{1}{x}\), both in the first quadrant.
Next, find the points of intersection between the two curves by setting \(x = \frac{5}{2} - \frac{1}{x}\) and solving for \(x\). Multiply both sides by \(x\) to clear the denominator, resulting in a quadratic equation.
Solve the quadratic equation to find the \(x\)-coordinates of the intersection points. Since the region is in the first quadrant, only consider positive solutions for \(x\).
Set up the integral for the area between the curves from the smaller intersection point \(x = a\) to the larger intersection point \(x = b\). The area is given by \(\int_a^b \left( \left( \frac{5}{2} - \frac{1}{x} \right) - x \right) \, dx\).
Finally, evaluate the integral by integrating each term separately and then subtracting the values at the bounds \(a\) and \(b\) to find the area of the region.

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Key Concepts

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

Finding Points of Intersection

To determine the area between two curves, first find where they intersect by setting the functions equal to each other. Solving y = 5/2 - 1/x and y = x gives the limits of integration, which define the region in the first quadrant.
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Critical Points

Setting Up the Definite Integral for Area

The area between two curves y = f(x) and y = g(x) over an interval [a, b] is found by integrating the difference f(x) - g(x). Here, identify which function is on top in the first quadrant and integrate their difference between the intersection points.
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Definition of the Definite Integral

Properties of Functions in the First Quadrant

Since the region is in the first quadrant, both x and y are positive. This restricts the domain and range for the problem, ensuring the limits of integration and the functions are considered only where x > 0 and y > 0.
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Properties of Functions
Related Practice
Textbook Question

53–62. Choose your method Let R be the region bounded by the following curves. Use the method of your choice to find the volume of the solid generated when R is revolved about the given axis.

y = x³,y=0, and x=2; about the x-axis

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Textbook Question

Suppose f and g have continuous derivatives on an interval [a, b]. Prove that if f(a)=g(a) and f(b)=g(b), then ∫a^b f′(x) dx = ∫a^b g′(x) dx.

Textbook Question

Find the area of the surface generated when the given curve is revolved about the given axis.


y=8√x, for 9≤x≤20; about the x-axis

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