What is the integral from 0 to 1 of (10y)/(e^{2y)} ?

The integral from 0 to 1 of (10y)/(e^{2y)} is 10(-3/(4e^2)+1/4)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral from 0 to 1 of (10y)/(e^{2y)}

\int_{0}^{1} \frac{10 y}{e ^{2 y}} d y

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

Decimal

1.48498 \dots

Solution steps

\int_{0}^{1} \frac{10 y}{e ^{2 y}} d y

Take the constant out:

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

= 10 \cdot \int_{0}^{1} \frac{y}{e ^{2 y}} d y

Apply Integration By Parts

= 10 [- \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}

= 10 [- \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}

= 10 (- \frac{3}{4 e ^{2}} + \frac{1}{4})

What is the integral from 0 to 1 of (10y)/(e^{2y)} ?
The integral from 0 to 1 of (10y)/(e^{2y)} is 10(-3/(4e^2)+1/4)