What is the integral from 1 to infinity of xe^{-15x} ?

The integral from 1 to infinity of xe^{-15x} is (16)/(225e^{15)}

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 xe^{-15x}

\int_{1}^{\infty} x e^{- 15 x} d x

\frac{16}{225 e ^{15}}

Decimal Notation

2.17531 E - 8

Solution steps

\int_{1}^{\infty} x e^{- 15 x} d x

Compute the indefinite integral:

\int x e^{- 15 x} d x = \frac{1}{225} (- 15 e^{- 15 x} x - e^{- 15 x}) + C

Compute the boundaries:

\int_{1}^{\infty} x e^{- 15 x} d x = 0 - (- \frac{16}{225 e ^{15}})

= 0 - (- \frac{16}{225 e ^{15}})

Simplify

= \frac{16}{225 e ^{15}}

\int_{0}^{2} x^{3} \cdot e^{x^{4}} d x

\int_{1}^{5} (x - \frac{5}{x})^{2} d x

\int_{0}^{\frac{π}{6}} (cos (3 x))^{8} d x

\int_{0}^{\frac{π}{6}} sin^{4} (x) d x

\int_{0}^{1} (2^{x} - 2) d x

What is the integral from 1 to infinity of xe^{-15x} ?
The integral from 1 to infinity of xe^{-15x} is (16)/(225e^{15)}