/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 52 Find the area of the region. $... [FREE SOLUTION] | 91影视

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Chapter 4: Problem 52

Find the area of the region. $$ y=\frac{e^{x}}{1+e^{2 x}} $$

Short Answer

Expert verified
The area under the curve \(y=\frac{e^{x}}{1+e^{2 x}}\) from \(a\) to \(b\) is given by \( \arctan(e^b) - \arctan(e^a)\).

Step by step solution

01

Identify the function

The function whose area under the curve needs to be found is given as \(y=\frac{e^{x}}{1+e^{2 x}}\). This function should be integrated to find the area under the curve.
02

Simplify the function

The function can be simplified by substituting \(e^{x}\) as \(t\). This substitution will also transform \(\mathrm{d}x\) into \(\mathrm{d}t / t\). So, the integral now becomes:\[\int \frac{t}{1+t^2} \frac{\mathrm{d}t}{t} = \int \frac{\mathrm{d}t}{1+t^2}\]This is a standard form.
03

Integrate

Now, the function can be integrated. The integral of the function \(\frac{1}{1+t^2}\) is known and is given by \( \arctan(t)\). Hence, the integral becomes:\[\int \frac{\mathrm{d}t}{1+t^2} = \arctan(t) = \arctan(e^x)\]This is the antiderivative of the function. To find the area under the curve from \(a\) to \(b\), subtract the values of the antiderivative at \(a\) and \(b\).
04

Find the area under the curve

The area under the curve is obtained by evaluating the antiderivative at the upper and lower limits, say \(a\) and \(b\), of the region of interest:\[Area = \arctan(e^b) - \arctan(e^a)\]

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