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Ch. 5 - Integration
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
All textbooksBriggs 3rd EditionCh. 5 - IntegrationProblem 5.1.73b
Chapter 5, Problem 5.1.73b

Mass from density A thin 10-cm rod is made of an alloy whose density varies along its length according to the function shown in the figure. Assume density is measured in units of g/cm. In Chapter 6, we show that the mass of the rod is the area under the density curve.
(b) Find the mass of the right half of the rod (5 ≤ x ≤ 10) .
Graph showing density in g/cm along a 10-cm rod, with varying density values plotted against length in cm.

Verified step by step guidance
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Step 1: Understand the problem. The mass of the rod is the area under the density curve. We are tasked with finding the mass of the right half of the rod, which corresponds to the interval 5 ≤ x ≤ 10.
Step 2: Analyze the graph. The density function is piecewise linear. From x = 5 to x = 6, the density increases linearly from 5 g/cm to 6 g/cm. From x = 6 to x = 10, the density remains constant at 6 g/cm.
Step 3: Break the interval into two subintervals: (a) x = 5 to x = 6, where the density is a linear function, and (b) x = 6 to x = 10, where the density is constant. Calculate the area under the curve for each subinterval separately.
Step 4: For the interval x = 5 to x = 6, the area is the area of a trapezoid. Use the formula for the area of a trapezoid: A = (1/2) * (base1 + base2) * height. Here, base1 = 5, base2 = 6, and height = 1 (difference in x-values).
Step 5: For the interval x = 6 to x = 10, the area is the area of a rectangle. Use the formula for the area of a rectangle: A = width * height. Here, width = 4 (difference in x-values) and height = 6 (density value). Add the areas from both subintervals to find the total mass of the right half of the rod.

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Key Concepts

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

Density Function

A density function describes how mass is distributed along a given length. In this case, the density of the rod varies with its length, represented graphically. Understanding the density function is crucial for calculating the mass, as it provides the necessary values to integrate over the specified interval.
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Properties of Functions

Integration

Integration is a fundamental concept in calculus used to find the area under a curve. In this context, the mass of the rod can be determined by integrating the density function over the specified interval (5 ≤ x ≤ 10). This process allows us to sum up the infinitesimal contributions of mass along the length of the rod.
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Integration by Parts for Definite Integrals

Definite Integral

A definite integral calculates the accumulation of quantities over a specific interval. For this problem, it involves evaluating the integral of the density function from x = 5 to x = 10. The result gives the total mass of the right half of the rod, reflecting the varying density across that segment.
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Definition of the Definite Integral
Related Practice
Textbook Question

Working with area functions Consider the function ƒ and the points a, b, and c.

(b) Graph ƒ and A.

ƒ(𝓍) = eˣ ; a = 0 , b = ln 2 , c = ln 4

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Textbook Question

Area functions for linear functions Consider the following functions ƒ and real numbers a (see figure).


b) Verify that A'(𝓍) = ƒ(𝓍).



ƒ(t) = 4t + 2 , a = 0

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The following functions are positive and negative on the given interval.

ƒ(𝓍) = xe⁻ˣ on [-1,1]

(b) Approximate the net area bounded by the graph of f and the x-axis on the interval using a left, right, and midpoint Riemann sum with n = 4.

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Textbook Question

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

                                                                                                                                                                                     (b) Suppose ƒ is a negative increasing function, for 𝓍 > 0 . Then the area function A(𝓍) = ∫₀ˣ ƒ(t) dt is a decreasing function of 𝓍 .

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Textbook Question

{Use of Tech} Riemann sums for larger values of n Complete the following steps for the given function f and interval.


ƒ(𝓍) = x² ― 1 on [2,5] ; n = 75

(b) Based on the approximations found in part (a), estimate the area of the region bounded by the graph of f and the x-axis on the interval.

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Textbook Question

{Use of Tech} Approximating net area The following functions are positive and negative on the given interval.

f(𝓍) = x³ on [-1,2]

(b) Approximate the net area bounded by the graph of f and the x-axis on the interval using a left, right, and midpoint Riemann sum with n = 4.

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