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Chapter 12: Problem 39

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. $$ y=\frac{x}{1+e^{x^{2}}}, y=0, x=2 $$

Short Answer

Expert verified
The exact area under the curve is 2. The graphing utility also shows an area close to 2, confirming our calculations.

Step by step solution

01

Evaluate the Definite Integral

The area A between the x-axis (y=0) and the curve \(y=\frac{x}{1+e^{x^{2}}}\) from \(x=0\) to \(x=2\) is given by: \[A = \int_0^2 \frac{x}{1+e^{x^{2}}} dx\]To evaluate this integral, it might be useful to make a variable substitution due to the presence of the exponential term in the denominator. Let \(u = x^{2}\), then \(du = 2x dx\), and \(x dx = \frac{1}{2} du\). The limits of integration will also change accordingly: for \(x=0\) we have \(u=0\), and for \(x=2\) we have \(u=4\). Therefore, the integral becomes: \[A = \frac{1}{2} \int_0^4 \frac{1}{1+e^{u}} du\]This is a simple integral to solve using direct integration techniques.
02

Solve the Integral

The integral in step one is a standard integral of the form \( \int \frac{1}{a+e^x} dx = ln|a+e^x| + C \). Applying this property:\[A = \frac{1}{2} ln|1+e^{u}| \Bigg|_0^4\]Simplifying this expression by substituting the limits back in should give us the exact area under the curve.
03

Evaluate the Area

Substituting the limits of integration into our result from step 2, we find:\[A = \frac{1}{2} (ln(1+e^{4}) - ln(1+e^{0}))= \frac{1}{2} ln(e^{4}) = 2 \]
04

Approximate Area Using Graphing Utility

To approximate the area under the curve, one can use a graphing utility. Graph the function \(y=\frac{x}{1+e^{x^{2}}}\) from \(x=0\) to \(x=2\) and use the tool's 'Area under Curve' function to get the area. The result should be very close to 2, which confirms the calculation in steps 2 and 3.

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Most popular questions from this chapter

Probability The probability of finding between \(a\) and \(b\) percent iron in ore samples is modeled by \(P(a \leq x \leq b)=\int_{a}^{b} 2 x^{3} e^{x^{2}} d x, \quad 0 \leq a \leq b \leq 1\) (see figure). Find the probabilities that a sample will contain between (a) \(0 \%\) and \(25 \%\) and (b) \(50 \%\) and \(100 \%\) iron.

Determine whether the improper integral diverges or converges. Evaluate the integral if it converges, and check your results with the results obtained by using the integration capabilities of a graphing utility. $$ \int_{0}^{1} \frac{1}{1-x} d x $$

Determine whether the improper integral diverges or converges. Evaluate the integral if it converges. $$ \int_{1}^{\infty} \frac{1}{\sqrt[3]{x}} d x $$

Approximate the integral using (a) the Trapezoidal Rule and (b) Simpson's Rule for the indicated value of \(n\). (Round your answers to three significant digits.) $$ \int_{0}^{2} \frac{1}{\sqrt{1+x^{3}}} d x, n=4 $$

Marginal Analysis In Exercises 27 and 28, use a program similar to the Simpson's Rule program on page 906 with \(n=4\) to approximate the change in revenue from the marginal revenue function \(d R / d x .\) In each case, assume that the number of units sold \(x\) increases from 14 to 16 . $$ \frac{d R}{d x}=5 \sqrt{8000-x^{3}} $$
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