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Chapter 5: Problem 20

Sketch the region bounded by the graphs of the algebraic functions and find the area of the region. $$ f(x)=\sqrt[3]{x-1}, g(x)=x-1 $$

Short Answer

Expert verified
The area under the curve is undefined as it is infinite.

Step by step solution

01

Sketch the Functions

First, sketch both functions \( f(x)=\sqrt[3]{x-1}\) and \(g(x)=x-1\) on the same graph for an initial visual understanding of the region. The function \( f(x)=\sqrt[3]{x-1} \) is a cubic root function shifted 1 unit to the right, and \( g(x)=x-1 \) is a straight line with a slope of 1.
02

Find the Intersection Points

Set \(f(x) = g(x)\) to find the points where the graphs intersect. These points will be the limits of integration when finding the area. Thus, by setting \(\sqrt[3]{x-1} =x-1\) and solving the equation, obtain \(x=1\) as the point of intersection.
03

Set Up the Integral for Area Calculation

Now, set up the integral for calculating the area between the two curves. The formula for the area between two curves from a to b is \( \int_a^b |f(x) - g(x)| dx\). Since the number of intersection points is only one, another value for 'b' in the integral can't be found. But by observing the graphs, it can be concluded that \(g(x)=x-1\) overtakes \(f(x)=\sqrt[3]{x-1}\) as \( x \to +\infty \). Thus the integral setting for the calculation becomes: \( \int_1^{+\infty}|x-1 - \sqrt[3]{x-1}| dx \)
04

Calculate the Area

Finally, compute the integral, an improper one, to find the area bounded by these two curves. However, this integral doesn't converge to a finite area. Therefore, the concept of the 'area' between these two curves doesn't exist in the usual sense as the area under the curve is infinite.

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Most popular questions from this chapter

Find the area of the region by integrating (a) with respect to \(x\) and (b) with respect to \(y\). $$ \begin{array}{l} y=x^{2} \\ y=6-x \end{array} $$

Use the integration capabilities of a graphing utility to approximate the volume of the solid generated by revolving the region bounded by the graphs of the equations about the \(x\) -axis. $$ y=2 \arctan (0.2 x), \quad y=0, \quad x=0, \quad x=5 $$

Sketch the region bounded by the graphs of the algebraic functions and find the area of the region. $$ y=x, \quad y=2-x, \quad y=0 $$

Let \(R\) be the region bounded by \(y=1 / x,\) the \(x\) -axis, \(x=1,\) and \(x=b,\) where \(b>1 .\) Let \(D\) be the solid formed when \(R\) is revolved about the \(x\) -axis. (a) Find the volume \(V\) of \(D\). (b) Write the surface area \(S\) as an integral. (c) Show that \(V\) approaches a finite limit as \(b \rightarrow \infty\). (d) Show that \(S \rightarrow \infty\) as \(b \rightarrow \infty\).

Writing Read the article "Arc Length, Area and the Arcsine Function" by Andrew M. Rockett in Mathematics Magazine. Then write a paragraph explaining how the arcsine function can be defined in terms of an arc length. (To view this article, go to the website www.matharticles.com.)
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