What is the integral from 0 to 1 of y/(e^{3y)} ?

The integral from 0 to 1 of y/(e^{3y)} is -4/(9e^3)+1/9

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to 1 of y/(e^{3y)}

\int_{0}^{1} \frac{y}{e ^{3 y}} d y

- \frac{4}{9 e ^{3}} + \frac{1}{9}

Decimal

0.08898 \dots

Solution steps

\int_{0}^{1} \frac{y}{e ^{3 y}} d y

Apply Integration By Parts

: [- \frac{1}{3} e^{- 3 y} y - \int - \frac{1}{3} e^{- 3 y} d y]_{0}^{1}

= [- \frac{1}{3} e^{- 3 y} y - \int - \frac{1}{3} e^{- 3 y} d y]_{0}^{1}

\int - \frac{1}{3} e^{- 3 y} d y = \frac{1}{9} e^{- 3 y}

= [- \frac{1}{3} e^{- 3 y} y - \frac{1}{9} e^{- 3 y}]_{0}^{1}

Compute the boundaries:

- \frac{4}{9 e ^{3}} + \frac{1}{9}

= - \frac{4}{9 e ^{3}} + \frac{1}{9}

\frac{d y}{d x} = x + 2 y

\int e^{6} θ sin (7 θ) d θ

\frac{d}{d y} (3 y^{2})

\frac{d}{d x} ((cos^{2} (x)) d x)

x \to 0 - lim (- \frac{1}{x ^{3}})

What is the integral from 0 to 1 of y/(e^{3y)} ?
The integral from 0 to 1 of y/(e^{3y)} is -4/(9e^3)+1/9