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Chapter 12: Problem 38

Use the table of integrals to find the exact area of the region bounded by the graphs of the equations. Then use a graphing utility to graph the region and approximate the area. $$ y=\frac{2}{1+e^{4 x}}, y=0, x=0, x=1 $$

Short Answer

Expert verified
The exact area of the region bounded by the given function and the x-axis between x=0 and x=1 is \( \frac{1-e^{-4}}{2} \)

Step by step solution

01

Function Analysis

Given the function \( y=\frac{2}{1+e^{4 x}} \) and the limits \( y=0, x=0,\) and \( x=1 \), the integral of y from x=0 to x=1 needs to be found. This will give the exact area under the graph between these points.
02

Integration of the Function

To find the integral, the exponential integral formula needs to be utilized, which states that the integral of \( e^ax \) dx = \( \frac{e^ax}{a} \). In this case, the fraction with exponential in the denominator can be rewritten as \( 2e^{-4x} \). Hence, \[ \int_{0}^{1} y dx = \int_{0}^{1} 2e^{-4x} dx \] which, using above formula, results in \( -\frac{2}{4}e^{-4x} \big|_0^1 \)
03

Evaluation of the Integral

Evaluating the integral expression at the upper and lower limits, the result is \[ -\frac{e^{-4}}{2} - \left(-\frac{1}{2}\right) = -\frac{e^{-4}-1}{2} \]
04

Solving for the Area

The region is bounded by y=0 and the x-axis, thus, the integral represents the exact area of this region. Even if the result was a negative value, the area is a scalar quantity and is always considered positive. Hence, the exact area is \[ A = \left| -\frac{e^{-4}-1}{2} \right| = \frac{1-e^{-4}}{2} \]

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