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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.R.41
Chapter 6, Problem 6.R.41

An area function Consider the functions y = x²/a and y = √x/a, where a>0. Find A(a), the area of the region between the curves.

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Identify the two functions given: \( y = \frac{x^2}{a} \) and \( y = \frac{\sqrt{x}}{a} \), where \( a > 0 \).
Determine the points of intersection by setting the two functions equal: \( \frac{x^2}{a} = \frac{\sqrt{x}}{a} \). Simplify and solve for \( x \) to find the limits of integration.
Set up the integral for the area \( A(a) \) between the curves. Since \( y = \frac{\sqrt{x}}{a} \) is above \( y = \frac{x^2}{a} \) in the interval between the points of intersection, the area is given by the integral \( A(a) = \int_{x_1}^{x_2} \left( \frac{\sqrt{x}}{a} - \frac{x^2}{a} \right) \, dx \), where \( x_1 \) and \( x_2 \) are the intersection points.
Factor out \( \frac{1}{a} \) from the integral to simplify: \( A(a) = \frac{1}{a} \int_{x_1}^{x_2} \left( \sqrt{x} - x^2 \right) \, dx \).
Evaluate the integral by integrating each term separately: \( \int \sqrt{x} \, dx = \int x^{1/2} \, dx \) and \( \int x^2 \, dx \). Then substitute the limits \( x_1 \) and \( x_2 \) to express \( A(a) \) in terms of \( a \).

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Key Concepts

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

Definite Integrals and Area Between Curves

The area between two curves over an interval is found by integrating the difference of the functions. Specifically, the integral of the upper function minus the lower function with respect to x gives the area between them. This requires identifying the correct limits of integration where the curves intersect.
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Finding Area Between Curves on a Given Interval

Finding Points of Intersection

To determine the limits of integration, solve for x where the two functions are equal. These intersection points define the interval over which the area between the curves is calculated. Setting y = x²/a equal to y = √x/a and solving for x is essential.
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Critical Points

Handling Parameters in Functions

The parameter 'a' affects the shape and position of the curves. When finding the area A(a), it is important to treat 'a' as a positive constant throughout the integration and algebraic manipulation. This ensures the final expression for area is correctly expressed in terms of 'a'.
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Eliminating the Parameter
Related Practice
Textbook Question

Comparing volumes Let R be the region bounded by y=1/x^p and the x-axis on the interval [1, a], where p>0 and a>1 (see figure). Let Vₓ and Vᵧ be the volumes of the solids generated when R is revolved about the x- and y-axes, respectively.


d. Find a general expression for Vᵧ in terms of a and p. Note that p=2 is a special case. What is Vᵧ when p=2?

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Textbook Question

Spring work


b. It takes 50 N of force to stretch a spring 0.2 m from its equilibrium position. How much work is needed to stretch it an additional 0.5 m?

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Textbook Question

Two methods The region R in the first quadrant bounded by the parabola y = 4-x² and coordinate axes is revolved about the y-axis to produce a dome-shaped solid. Find the volume of the solid in the following ways:


b. Apply the shell method and integrate with respect to x.

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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Textbook Question

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

Textbook Question

An oscillator The acceleration of an object moving along a line is given by a(t) = 2 sin πt/4. The initial velocity and position are v(0)= −8/π and s(0)=0.

 a. Find the velocity and position for t≥0.