Chapter 7: Problem 47
As \((x+b)^{2}+C\) or \(-(x-b)^{2}+C\) by completing the square. $$ x^{2}+2 x+1 $$
Short Answer
Step by step solution
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
/*! 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}
Learning Materials
Features
Discover
Chapter 7: Problem 47
As \((x+b)^{2}+C\) or \(-(x-b)^{2}+C\) by completing the square. $$ x^{2}+2 x+1 $$
These are the key concepts you need to understand to accurately answer the question.
All the tools & learning materials you need for study success - in one app.
Get started for free
How would you compute \(\int x e^{x} \tan ^{-1} x d x\) ? Don't do it.
As \((x+b)^{2}+C\) or \(-(x-b)^{2}+C\) by completing the square. $$ -x^{2}+10 $$
In \(\delta(x-4)\), the spike shifts to \(x=\frac{1}{2}\). It is the derivative of the shifted step \(U\left(x-\frac{1}{2}\right)\). The integral of \(v(x) \delta\left(x-\frac{1}{2}\right)\) equals the value of \(v\) at \(x=\frac{1}{2}\). Compute (a) \(\int_{0}^{1} \delta\left(x-\frac{1}{2}\right) d x\) (b) \(\int_{0}^{1} e^{x} \delta\left(x-\frac{1}{2}\right) d x\) (c) \(\int_{-1}^{1} \delta(x) \delta\left(x-\frac{1}{2}\right) d x\)
Integrate by substitution. Change \(\theta\) back to \(x\). $$ \int \frac{x^{3} d x}{\sqrt{9-x^{2}}} $$
For which \(p\) is \(\int_{0}^{\infty} \frac{d x}{x^{p}+x^{-p}}=\infty ?\)
What do you think about this solution?
We value your feedback to improve our textbook solutions.