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Chapter 5: Problem 19

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 of the region between the graphs of \(\sqrt{3x}+1\) and \(x+1\) is infinite.

Step by step solution

01

Identify the Intersections

Set the functions equal to each other to find where they intersect, i.e., solve \(\sqrt{3x} + 1 = x + 1\) for x. After simplifying this equation, find \(x = 0\).
02

Understand the Region

The graphs of the functions \(y = \sqrt{3x} + 1\) and \(y = x + 1\) intersect at \(x = 0\) and form a region between them on the interval \([0,\infty)\). The function \(y = \sqrt{3x} + 1\) is above the line \(y = x + 1\) on this interval.
03

Calculate the Area between the Graphs

The area A between two curves from \(x = a\) to \(x = b\) is defined by the definite integral \(A = \int_a^b |f(x) - g(x)| dx\). For this case, the area is given by the definite integral \(A = \int_0^\infty ((\sqrt{3x}+1) - (x+1)) dx\) = \(\int_0^\infty (\sqrt{3x} - x) dx\).
04

Compute the Integral

The integral \(\int_0^\infty (\sqrt{3x} - x) dx\) diverges as it is over an infinite range and the function being integrated is not bounded. The area between these two graphs is infinite. Interpreting this result, the region formed by these two graphs continues infinitely to the right.

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