What is the integral from 0 to 1 of (3y)/(e^{2y)} ?

The integral from 0 to 1 of (3y)/(e^{2y)} is 3(-3/(4e^2)+1/4)

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 (3y)/(e^{2y)}

\int_{0}^{1} \frac{3 y}{e ^{2 y}} d y

3 (- \frac{3}{4 e ^{2}} + \frac{1}{4})

Decimal

0.44549 \dots

Solution steps

\int_{0}^{1} \frac{3 y}{e ^{2 y}} d y

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 3 \cdot \int_{0}^{1} \frac{y}{e ^{2 y}} d y

Apply Integration By Parts

: [- \frac{1}{2} e^{- 2 y} y - \int - \frac{1}{2} e^{- 2 y} d y]_{0}^{1}

= 3 [- \frac{1}{2} e^{- 2 y} y - \int - \frac{1}{2} e^{- 2 y} d y]_{0}^{1}

\int - \frac{1}{2} e^{- 2 y} d y = \frac{1}{4} e^{- 2 y}

= 3 [- \frac{1}{2} e^{- 2 y} y - \frac{1}{4} e^{- 2 y}]_{0}^{1}

Compute the boundaries:

- \frac{3}{4 e ^{2}} + \frac{1}{4}

= 3 (- \frac{3}{4 e ^{2}} + \frac{1}{4})

\int ln (2 x + 2) d x

\frac{d}{d x} (x^{2} - 6 x + 7)

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

\int 17 tan^{5} (x) sec^{4} (x) d x

\int x^{- 1.1} d x

What is the integral from 0 to 1 of (3y)/(e^{2y)} ?
The integral from 0 to 1 of (3y)/(e^{2y)} is 3(-3/(4e^2)+1/4)