 View inside, the captions and chapters, a quick preview of the content, the table of contents, and links to other parts of the book .

Category:2003 non-fiction books
Category:Electronics engineering books
Category:Engineering mathematics
Category:Products and services discontinued in 2008
Category:Networking booksQ:

$\int_0^\infty \sqrt{x}\, dx =\int_0^\infty \sqrt{y}\, dy$?

I was doing some questions from an exam, and as i don’t have the books, I need to check my answers on internet, And i found this problem.
I need to check it, please tell me if the identity is true or not

A:

There is a solution you do not try, so it is of no use to you, but nobody’s presence is needed to prove that it is a real function.
So, we have $f(x) = g(x)$ and $g(x) = f(x)$. By the uniqueness property of integrals, we have $f(x) = g(x) = \int_a^x f(t) \, dt = \int_a^x g(t) \, dt = \int_a^x f(t) \, dt = g(x) = f(x)$.
See why this identity is true here. And see also how it works when $a = 0$.
Another proof that the identity is true is that there is a $c$ in the interval $[0,\infty)$ such that $f(c) = g(c)$.
It can be seen here that $f(c) = g(c) = \int_0^c \sqrt{t} \, dt = \int_c^\infty \sqrt{t} \, dt$
This also shows that your identity is false.

Landmark Sites

Rigveda Compendium and Sama Veda

The Rigveda Compendium and Sama Veda, (from Sanskrit: रिवेद संजय) comprise the literature on the

Tags: K. Solutions Pdf Free Download Pdf Electrical Networks , Electrical Networks, Electrical Engineering, Electronics, Electronics&Electrical Engineering, Electrical Circuit Theory, Advanced Circuit Analysis, Economics, Electrical And Instrumentation Engineering, Electrical