Problem 19 - Essential Calculus

Question

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

Step-by-Step Solution

Answer

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

Step 1: 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 \(x = 0\).

Step 2: 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.

Step 3: 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 \(A = \int_0^\infty ((\sqrt{3x}+1) - (x+1)) dx\) = \(\int_0^\infty (\sqrt{3x} - x) dx\).

Step 4: 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.