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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 6.2.49
Chapter 6, Problem 6.2.49

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


The region in the first quadrant bounded by y=x^2/3 and y=4

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Step 1: Identify the region to be calculated. The region is in the first quadrant, bounded by the curve y = x^(2/3) and the horizontal line y = 4. The intersection points of these two curves will determine the limits of integration.
Step 2: Solve for the intersection points by equating y = x^(2/3) and y = 4. This means solving the equation x^(2/3) = 4 for x. Rewrite the equation as x = 4^(3/2) to find the upper limit of integration.
Step 3: Set up the integral to calculate the area. The area is given by the integral of the difference between the upper curve (y = 4) and the lower curve (y = x^(2/3) over the interval [0, 4^(3/2)]. The integral is: ∫[0, 4^(3/2)] (4 - x^(2/3)) dx.
Step 4: Break the integral into two parts for clarity: ∫[0, 4^(3/2)] 4 dx - ∫[0, 4^(3/2)] x^(2/3) dx. Compute each integral separately. For the first integral, ∫[0, 4^(3/2)] 4 dx, use the formula for the integral of a constant. For the second integral, ∫[0, 4^(3/2)] x^(2/3) dx, use the power rule for integration.
Step 5: Combine the results of the two integrals to find the total area. Simplify the expressions obtained from the integration process to express the area in its final form.

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Key Concepts

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

Definite Integral

A definite integral calculates the area under a curve between two specified points on the x-axis. It is represented as ∫[a, b] f(x) dx, where f(x) is the function being integrated, and a and b are the limits of integration. This concept is essential for finding the area of regions bounded by curves.
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Definition of the Definite Integral

Area Between Curves

The area between two curves can be found by integrating the difference of the functions that define the curves over a specified interval. If f(x) is the upper curve and g(x) is the lower curve, the area A is given by A = ∫[a, b] (f(x) - g(x)) dx. This concept is crucial for solving the given problem, as it involves finding the area between y = x^(2/3) and y = 4.
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Finding Area Between Curves on a Given Interval

Finding Intersection Points

To determine the area between two curves, it is necessary to find their points of intersection, as these points will serve as the limits of integration. This involves solving the equation where the two functions are equal, which provides the x-values that define the region of interest. In this case, solving y = x^(2/3) and y = 4 will yield the limits for the definite integral.
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Critical Points
Related Practice
Textbook Question

52–54. Force on a window A diving pool that is 4 m deep and full of water has a viewing window on one of its vertical walls. Find the force on the following windows. 


The window is circular, with a radius of 0.5 m, tangent to the bottom of the pool.

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Textbook Question

39–44. Shell method about other lines Let R be the region bounded by y = x²,x=1, and y=0. Use the shell method to find the volume of the solid generated when R is revolved about the following lines.


x =2

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Textbook Question

13–20. Mass of one-dimensional objects Find the mass of the following thin bars with the given density function.


ρ(x) = {1 if 0≤x≤2 {2 if 2<x≤3

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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.

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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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Textbook Question

Let f(x) = {x   if 0≤x≤2

      2x−2  if 2<x≤5

      −2x+18 if 5<x≤6. 


Find the volume of the solid formed when the region bounded by the graph of f, the x-axis, and the line x=6 is revolved about the x-axis.

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