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Chapter 6: Problem 38

Find the area of the region bounded by the graphs of the equations. $$ y=\frac{x}{1+e^{x^{2}}}, y=0, x=2 $$

Short Answer

Expert verified
The area is approximated by the integral \[A=\int_{{0}}^{{2}}\frac{x}{1+e^{x^{2}}}~dx\] It requires numerical methods to be solved.

Step by step solution

01

Understanding the problem

The problem provides the equations \(y=\frac{x}{1+e^{x^{2}}}\), \(y=0\), and \(x=2\). The graph of the function \(y=\frac{x}{1+e^{x^{2}}}\) lies above the x-axis between \(x=0\) and \(x=2\), while \(y=0\) represents the x-axis. So we are asked to find the area of the region between this curve and the x-axis, up to \(x=2\).
02

Setting up the integral

To find the area bounded by the curve and the x-axis from \(x=0\) to \(x=2\), we will use the area formula: \[A=\int_{{a}}^{{b}}|f(x)|~dx\], where \(a\) and \(b\) are the points where the graph cuts the x-axis. In this case, \(a=0\), \(b=2\), and \(f(x)=\frac{x}{1+e^{x^{2}}}\)
03

Calculating the area

Substituting the limits and the function into the area formula, we get: \[A=\int_{{0}}^{{2}}\left|\frac{x}{1+e^{x^{2}}}\right|~dx\]Since the function is positive in the range \(x=0\) to \(x=2\), we can remove the absolute signs. Now, the function to integrate has become: \[A=\int_{{0}}^{{2}}\frac{x}{1+e^{x^{2}}}~dx\] Since this integral is non elementary, we can use a numerical method like the Simpson's rule or a power series to approximate the area.

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Most popular questions from this chapter

(a) Let \(f^{\prime}(x)\) be continuous. Show that \(\lim _{h \rightarrow 0} \frac{f(x+h)-f(x-h)}{2 h}=f^{\prime}(x)\) (b) Explain the result of part (a) graphically.

In Exercises 65 and 66, apply the Extended Mean Value Theorem to the functions \(f\) and \(g\) on the given interval. Find all values \(c\) in the interval \((a, b)\) such that \(\frac{f^{\prime}(c)}{g^{\prime}(c)}=\frac{f(b)-f(a)}{g(b)-g(a)}\) \(\begin{array}{l} \underline{\text { Functions }} \\ f(x)=\ln x, \quad g(x)=x^{3} \end{array} \quad \frac{\text { Interval }}{\left[1,4\right]}\)

Sketch the graph of \(g(x)=\left\\{\begin{array}{ll}e^{-1 / x^{2}}, & x \neq 0 \\ 0, & x=0\end{array}\right.\) and determine \(g^{\prime}(0)\).

Determine all values of \(p\) for which the improper integral converges. $$ \int_{0}^{1} \frac{1}{x^{p}} d x $$

Show that \(\lim _{x \rightarrow \infty} \frac{x^{n}}{e^{x}}=0\) for any integer \(n>0\).
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