What is the integral from 2 to infinity of ye^{-6y} ?

The integral from 2 to infinity of ye^{-6y} is (13)/(36e^{12)}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 2 to infinity of ye^{-6y}

\int_{2}^{\infty} y e^{- 6 y} d y

\frac{13}{36 e ^{12}}

Solution steps

\int_{2}^{\infty} y e^{- 6 y} d y

Compute the indefinite integral:

\int y e^{- 6 y} d y = \frac{1}{36} (- 6 e^{- 6 y} y - e^{- 6 y}) + C

Compute the boundaries:

\int_{2}^{\infty} y e^{- 6 y} d y = 0 - (- \frac{13}{36 e ^{12}})

= 0 - (- \frac{13}{36 e ^{12}})

Simplify

= \frac{13}{36 e ^{12}}

y = \frac{π}{3} (l^{2} h - h^{3})

\frac{d}{d x} (\frac{1}{2 x ^{\frac{3}{2}}})

x^{3}, 2

(18 y^{2} - x y) d x + x^{2} d y = 0

f (x) = x^{2} - 1, (- 4, 15)

What is the integral from 2 to infinity of ye^{-6y} ?
The integral from 2 to infinity of ye^{-6y} is (13)/(36e^{12)}