What is the integral from 1 to infinity of xsin(2x) ?

The integral from 1 to infinity of xsin(2x) is diverges

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 1 to infinity of xsin(2x)

\int_{1}^{\infty} x sin (2 x) d x

diverges

Solution steps

\int_{1}^{\infty} x sin (2 x) d x

Compute the indefinite integral:

\int x sin (2 x) d x = - \frac{1}{2} x cos (2 x) + \frac{1}{4} sin (2 x) + C

Compute the boundaries:

\int_{1}^{\infty} x sin (2 x) d x =

diverges

= diverges

\int_{0}^{x^{2}} e^{t^{3}} d t

\int_{1}^{10} x^{2} (x - 1) d x

\int_{- 2}^{1} 3 x^{2} + 2 d x

\int_{3}^{2} x d x

\int_{- 1}^{2} (1 - x^{2}) d x

What is the integral from 1 to infinity of xsin(2x) ?
The integral from 1 to infinity of xsin(2x) is diverges