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Chapter 7: Problem 17

Evaluate the following integrals or state that they diverge. $$\int_{-\infty}^{\infty} x e^{-x^{2}} d x$$

Short Answer

Expert verified
#Answer#: The integral of the function $$f(x) = xe^{-x^{2}}$$ over the range $$(-\infty, \infty)$$ is equal to 0, as f(x) is an odd function and its integral over a symmetric interval will cancel out the positive and negative parts.

Step by step solution

01

Observing the Symmetry of the Function

The function has an interesting property: it is an odd function. An odd function satisfies the property $$f(-x) = -f(x)$$. In this case, $$f(-x) = (-x)e^{-(-x)^{2}} = - xe^{-x^{2}} = -f(x)$$ This symmetry property will be useful when we calculate the integral.
02

Evaluating the Integral of an Odd Function

When integrating an odd function over a symmetric interval, like $$(-\infty, \infty)$$, the integral will equal zero. This is because the positive and negative parts of the function will exactly cancel each other out. Symbolically, for an odd function, we have $$\int_{-\infty}^{\infty} f(x) dx = \int_{-\infty}^{0} f(x) dx + \int_{0}^{\infty} f(x) dx = \int_{0}^{\infty} (-f(x)) dx + \int_{0}^{\infty} f(x) dx = 0$$ Therefore, the integral of the given function is simply zero.

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Most popular questions from this chapter

Approximate the following integrals using Simpson's Rule. Experiment with values of \(n\) to ensure that the error is less than \(10^{-3}\). \(\int_{0}^{\pi} \sin 6 x \cos 3 x d x=\frac{4}{9}\)

Given a Midpoint Rule approximation \(M(n)\) and a Trapezoid Rule approximation \(T(n)\) for a continuous function on \([a, b]\) with \(n\) subintervals, show that \(T(2 n)=(T(n)+M(n)) / 2\).

A differential equation and its direction field are given. Sketch a graph of the solution that results with each initial condition. $$\begin{aligned}&y^{\prime}(t)=\frac{\sin t}{y},\\\&y(-2)=-2 \text { and }\\\&y(-2)=2\end{aligned}$$

Determine whether the following statements are true and give an explanation or counterexample. a. The general solution of \(y^{\prime}(t)=20 y\) is \(y=e^{20 t}\). b. The functions \(y=2 e^{-2 t}\) and \(y=10 e^{-2 t}\) do not both satisfy the differential equation \(y^{\prime}+2 y=0\). c. The equation \(y^{\prime}(t)=t y+2 y+2 t+4\) is not separable. d. A solution of \(y^{\prime}(t)=2 \sqrt{y}\) is \(y=(t+1)^{2}\).

A differential equation of the form \(y^{\prime}(t)=F(y)\) is said to be autonomous (the function \(F\) depends only on \(y\) ). The constant function \(y=y_{0}\) is an equilibrium solution of the equation provided \(F\left(y_{0}\right)=0\) (because then \(y^{\prime}(t)=0,\) and the solution remains constant for all \(t\) ). Note that equilibrium solutions correspond to horizontal line segments in the direction field. Note also that for autonomous equations, the direction field is independent of \(t\). Consider the following equations. a. Find all equilibrium solutions. b. Sketch the direction field on either side of the equilibrium solutions for \(t \geq 0\). c. Sketch the solution curve that corresponds to the initial condition \(y(0)=1\). $$y^{\prime}(t)=y(2-y)$$
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