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Chapter 6: Problem 15

Determine whether the improper integral diverges or converges. Evaluate the integral if it converges. $$ \int_{-\infty}^{0} x e^{-2 x} d x $$

Short Answer

Expert verified
The improper integral \(\int_{-\infty}^{0} x e^{-2x} dx\) converges and its value is \(\frac{1}{2}\).

Step by step solution

01

Swap the Limits

Swap the limits of integration to get rid of the negative infinity. This step results in a change of sign during integration. Therefore, our integral becomes: \(-\int_{0}^{\infty} x e^{-2x} dx\)
02

Integration by Parts

Utilize integration by parts theorem, let \(u=x\) and \(dv=e^{-2x} dx\). Calculating for \(du\) and \(v\), we get \(du = dx\) and \(v = -\frac{1}{2}e^{-2x}\). The integration by parts theorem gives us that the integral of \(u dv\) is equal to \(uv - \int v du\). Substituting for \(u\), \(v\), \(du\) and \(dv\) we get the equation \(-(-x\frac{1}{2}e^{-2x}-\frac{1}{2}\int_{0}^{\infty} e^{-2x} dx)\). This simplifies to \(x\frac{1}{2}e^{-2x}+\frac{1}{2}\int_{0}^{\infty} e^{-2x} dx\).
03

Solve the remaining integral

Solve the remaining integral, that is, \(\int_{0}^{\infty} e^{-2x} dx\). This is a standard integral and its solution is \(-\frac{1}{2}e^{-2x}\) evaluated from 0 to infinity which gives us \(\frac{1}{2}\).
04

Substitute values

Substitute the values of \(x\) and the evaluated integral solution into the equation from Step 2, giving us the following expression: \[\lim_{{x-> \infty}} [x\frac{1}{2}e^{-2x}]+\frac{1}{2}.\] The limit of \(x\frac{1}{2}e^{-2x}\) as \(x\) approaches infinity is 0. Therefore, our final answer becomes \(\frac{1}{2}\).

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Most popular questions from this chapter

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 \(g^{\prime}(0)\).

Evaluate the definite integral. $$ \int_{-\pi / 2}^{\pi / 2}\left(\sin ^{2} x+1\right) d x $$

True or False? In Exercises 67-70, determine whether the statement is true or false. If it is false, explain why or give an example that shows it is false. If \(p(x)\) is a polynomial, then \(\lim _{x \rightarrow \infty}\left[p(x) / e^{x}\right]=0\).

(a) Use a graphing utility to graph the function \(y=e^{-x^{2}}\). (b) Show that \(\int_{0}^{\infty} e^{-x^{2}} d x=\int_{0}^{1} \sqrt{-\ln y} d y\).

(a) The improper integrals \(\int_{1}^{\infty} \frac{1}{x} d x \quad\) and \(\int_{1}^{\infty} \frac{1}{x^{2}} d x\) diverge and converge, respectively. Describe the essential differences between the integrands that cause one integral to converge and the other to diverge. (b) Sketch a graph of the function \(y=\sin x / x\) over the interval \((1, \infty)\). Use your knowledge of the definite integral to make an inference as to whether or not the integral \(\int_{1}^{\infty} \frac{\sin x}{x} d x\) converges. Give reasons for your answer. (c) Use one iteration of integration by parts on the integral in part (b) to determine its divergence or convergence.
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