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Ch. 6 - Applications of Integration
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Non è quello che usi tu?Cambia libro di testo
Tutti i libri di testoBriggs 3rd EditionCh. 6 - Applications of IntegrationProblema 6.R.23
Capitolo 6, Problema 6.R.23

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


The region in the first quadrant bounded by y = x/6 and y = 1−|x/2−1|

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First, identify the curves that bound the region in the first quadrant: the line \(y = \frac{x}{6}\) and the piecewise function \(y = 1 - \left| \frac{x}{2} - 1 \right|\).
Next, analyze the function \(y = 1 - \left| \frac{x}{2} - 1 \right|\) by considering the expression inside the absolute value. Set \(\frac{x}{2} - 1 = 0\) to find the critical point \(x = 2\) where the function changes its form.
Rewrite \(y = 1 - \left| \frac{x}{2} - 1 \right|\) as a piecewise function: For \(x \leq 2\), \(y = 1 - (1 - \frac{x}{2}) = \frac{x}{2}\), For \(x > 2\), \(y = 1 - (\frac{x}{2} - 1) = 2 - \frac{x}{2}\).
Determine the points of intersection between \(y = \frac{x}{6}\) and each piece of the piecewise function: - For \(x \leq 2\), solve \(\frac{x}{6} = \frac{x}{2}\), - For \(x > 2\), solve \(\frac{x}{6} = 2 - \frac{x}{2}\). These intersection points will serve as limits of integration.
Set up the integral(s) for the area of the region in the first quadrant bounded by the curves. The area can be found by integrating the difference between the upper and lower functions over the appropriate intervals determined by the intersection points.

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Understanding Piecewise and Absolute Value Functions

The function y = 1 − |x/2 − 1| involves an absolute value, which creates a piecewise definition. Understanding how to rewrite and interpret absolute value expressions as piecewise linear functions is essential to identify the shape and boundaries of the region.
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Piecewise Functions

Finding Points of Intersection

Determining the area bounded by two curves requires finding their points of intersection. Solving y = x/6 and y = 1 − |x/2 − 1| simultaneously helps establish the limits of integration for calculating the area.
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Critical Points

Definite Integration to Find Area Between Curves

The area between two curves is found by integrating the difference of the upper and lower functions over the interval defined by their intersection points. Setting up and evaluating the definite integral accurately yields the desired area.
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Finding Area Between Curves on a Given Interval
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{Use of Tech} Decreasing velocity A projectile is fired upward, and its velocity in m/s is given by v(t) = 200e^−t/10, for t≥0.

a. Graph the velocity function, for t≥0.

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27–33. Multiple regions The regions R₁,R₂, and R₃ (see figure) are formed by the graphs of y = 2√x,y = 3−x,and x=3.

Find the area of each of the regions R₁,R₂, and R₃.

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Force on a dam Find the total force on the face of a semicircular dam with a radius of 20 m when its reservoir is full of water. The diameter of the semicircle is the top of the dam.

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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.


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27–33. Multiple regions The regions R₁,R₂, and R₃ (see figure) are formed by the graphs of y = 2√x,y = 3−x,and x=3.


Use the shell method to find an integral, or sum of integrals, that equals the volume of the solid obtained by revolving region R₃ about the line x=3. Do not evaluate the integral.

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43–55. Volumes of solids Choose the general slicing method, the disk/washer method, or the shell method to answer the following questions.


The region bounded by the graph of y = 4−x² and the x-axis on the interval [−2,2] is revolved about the line x = −2. What is the volume of the solid that is generated?

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