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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.34a
Chapter 6, Problem 6.R.34a

Area and volume The region R is bounded by the curves x = y²+2,y=x−4, and y=0 (see figure).
Graph showing region R bounded by the curves x = y² + 2, y = x - 4, and y = 0 on a coordinate plane.
a. Write a single integral that gives the area of R.

Verified step by step guidance
1
Identify the curves that bound the region R: the parabola \(x = y^{2} + 2\), the line \(y = x - 4\), and the line \(y = 0\) (the x-axis).
Express the line \(y = x - 4\) in terms of \(x\) to get \(x = y + 4\) by solving for \(x\).
Determine the interval for \(y\) over which the region exists by finding the points where the curves intersect and where \(y = 0\) bounds the region. Since \(y = 0\) is the lower boundary, find the \(y\)-values where \(x = y^{2} + 2\) and \(x = y + 4\) intersect.
Set up the integral for the area by integrating with respect to \(y\), where the integrand is the difference between the rightmost curve and the leftmost curve in terms of \(x\). This gives the integral: \(\int_{y=a}^{y=b} \left[(y + 4) - (y^{2} + 2)\right] \, dy\) where \(a\) and \(b\) are the limits found in the previous step.
Simplify the integrand to \(y + 4 - y^{2} - 2 = -y^{2} + y + 2\) and write the final integral expression for the area as \(\int_{a}^{b} (-y^{2} + y + 2) \, dy\).

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Key Concepts

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

Setting up integrals for area between curves

To find the area of a region bounded by curves, we integrate the difference between the functions that define the boundaries. When curves are given as functions of y or x, the integral limits correspond to the intersection points, and the integrand is the difference of the outer and inner functions along the axis of integration.
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Expressing curves in terms of a single variable

When curves are given in different forms, such as x in terms of y or y in terms of x, it is often necessary to rewrite all curves in terms of the same variable to set up the integral. In this problem, since one curve is x = y² + 2 and another is y = x - 4, rewriting y = x - 4 as x = y + 4 helps integrate with respect to y.
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Finding intersection points to determine limits of integration

The limits of integration are found by solving for the points where the bounding curves intersect. These points define the interval over which the area is calculated. For this problem, solving x = y² + 2 and x = y + 4 simultaneously gives the y-values that serve as integration limits.
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