Ch. 6 - Applications of Integration

Briggs - Calculus: Early Transcendentals 3rd Edition

Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook

Briggs 3rd Edition

Ch. 6 - Applications of Integration

Problem 6.2.1

Chapter 6, Problem 6.2.1

Set up a sum of two integrals that equals the area of the shaded region bounded by the graphs of the functions f and g on [a, c] (see figure). Assume the curves intersect at x=b.

Verified step by step guidance

Identify the points of intersection of the two functions, which are given as x = a, x = b, and x = c. The shaded region lies between these points.

Determine which function is on top and which is on the bottom in each subinterval. From the graph, on the interval [a, b], the function f(x) is above g(x), and on the interval [b, c], the function g(x) is above f(x).

Set up the integral for the area on the interval [a, b] as the integral of the difference between the top function and the bottom function: \(\int_{a}^{b} (f(x) - g(x)) \, dx\).

Set up the integral for the area on the interval [b, c] as the integral of the difference between the top function and the bottom function: \(\int_{b}^{c} (g(x) - f(x)) \, dx\).

Express the total shaded area as the sum of the two integrals: \(\int_{a}^{b} (f(x) - g(x)) \, dx + \int_{b}^{c} (g(x) - f(x)) \, dx\).

Verified video answer for a similar problem:

This video solution was recommended by our tutors as helpful for the problem above.

Video duration:

Was this helpful?

Key Concepts

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

Definite Integrals and Area Under a Curve

A definite integral calculates the net area between a function's graph and the x-axis over a specific interval. When the function lies above the x-axis, the integral represents the area directly; if below, it represents the negative area. Understanding this helps in setting up integrals to find areas bounded by curves.

Area Between Two Curves

The area between two curves f(x) and g(x) over an interval [a, c] is found by integrating the difference of the functions, |f(x) - g(x)|, over that interval. When the curves intersect at x = b, the integral must be split at b to account for which function is on top in each subinterval.

Splitting Integrals at Points of Intersection

When two curves intersect within the interval of interest, the relative position of the functions changes. To correctly compute the area between them, the integral is split at the intersection point, integrating the difference with the upper function minus the lower function on each subinterval separately.

Recommended video:

06:18

Integration by Parts for Definite Integrals

Related Practice

Textbook Question

Determine the area of the shaded region in the following figures.

Textbook Question

9–20. Arc length calculations Find the arc length of the following curves on the given interval.

y = (x²+2)^3/2 / 3 on [0, 1]

views

Textbook Question

9-34. Shell method Let R be the region bounded by the following curves. Use the shell method to find the volume of the solid generated when R is revolved about indicated axis.

y = 1−x²,x = 0, and y = 0, in the first quadrant; about the y-axis

views

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.

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+sin x, for 0≤x≤π

Textbook Question

Work from force How much work is required to move an object from x=0 to x=3 (measured in meters) in the presence of a force (in N) given by F(x)=2x acting along the x-axis?

views