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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 6.2.35a
Chapter 6, Problem 6.2.35a

For the given regions R₁ and R₂, complete the following steps.


a. Find the area of region R₁.


R₁is the region in the first quadrant bounded by the line x=1 and the curve y=6x(2−x^2)^2; R₂ is the region in the first quadrant bounded the curve y=6x(2−x^2)^2and the line y=6x.

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Identify the boundaries of region R₁. It is bounded by the line \(x=1\), the curve \(y=6x(2 - x^2)^2\), and the coordinate axes in the first quadrant. Since it is in the first quadrant, \(x\) ranges from 0 to 1.
Set up the integral for the area of region R₁. Because the region is bounded vertically by the curve \(y=6x(2 - x^2)^2\) and horizontally by \(x=1\), the area can be found by integrating the function \(y=6x(2 - x^2)^2\) with respect to \(x\) from 0 to 1:
\[\text{Area}_{R_1} = \int_0^1 6x(2 - x^2)^2 \, dx.\]
Before integrating, consider expanding the integrand \(6x(2 - x^2)^2\) to simplify the integral. First, expand \((2 - x^2)^2\) using the binomial formula:
\[(2 - x^2)^2 = 4 - 4x^2 + x^4.\]
Multiply this expansion by \$6x\( to get the integrand in polynomial form, then integrate term-by-term over the interval \)[0,1]$ to find the area of region R₁.

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Key Concepts

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

Definite Integrals for Area Calculation

Definite integrals are used to find the area under a curve between two points on the x-axis. By integrating the function that defines the curve over the given interval, we obtain the exact area of the region bounded by the curve and the x-axis or vertical lines.
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Setting Up Integration Limits from Geometric Boundaries

Determining the correct limits of integration is essential and depends on the boundaries of the region. For regions bounded by vertical lines and curves, the x-values where these boundaries intersect define the integration limits, ensuring the integral covers the entire region.
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Area Between Two Curves

When finding the area between two curves, subtract the lower function from the upper function and integrate the difference over the interval where they intersect. This method calculates the net area enclosed between the two curves accurately.
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Related Practice
Textbook Question

17–22. Position from velocity Consider an object moving along a line with the given velocity v and initial position.


a. Determine the position function, for t≥0, using the antiderivative method


v(t) = 9−t² on [0, 4]; s(0)=−2

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Textbook Question

Functions from arc length What differentiable functions have an arc length on the interval [a, b] given by the following integrals? Note that the answers are not unique. Give a family of functions that satisfy the conditions.

a. ∫a^b √1+16x⁴ dx

Textbook Question

Explain why or why not Determine whether the following statements are true and give an explanation or counterexample.


a. The distance traveled by an object moving along a line is the same as the displacement of the object.

Textbook Question

{Use of Tech} Oscillating motion A mass hanging from a spring is set in motion, and its ensuing velocity is given by v(t) = 2π cos πt, for t≥0. Assume the positive direction is upward and s(0)=0. 


a. Determine the position function, for t≥0.

Textbook Question

9–10. Velocity graphs The figures show velocity functions for motion along a line. Assume the motion begins with an initial position of s(0)=0. Determine the following.

a. The displacement between t=0 and t=5

Textbook Question

A surface is generated by revolving the line f(x)=2−x, for 0≤x≤2, about the x-axis. Find the area of the resulting surface in the following ways.


a. Using calculus