What is the integral from 0 to 5 of 2te^{-t^2} ?

The integral from 0 to 5 of 2te^{-t^2} is 1-1/(e^{25)}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to 5 of 2te^{-t^2}

\int_{0}^{5} 2 t e^{- t^{2}} d t

1 - \frac{1}{e ^{25}}

Decimal

1

Solution steps

\int_{0}^{5} 2 t e^{- t^{2}} d t

Take the constant out:

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

= 2 \cdot \int_{0}^{5} t e^{- t^{2}} d t

Apply u-substitution

: \int_{0}^{- 25} - \frac{e ^{u}}{2} d u

= 2 \cdot \int_{0}^{- 25} - \frac{e ^{u}}{2} d u

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= 2 (- \int_{- 25}^{0} - \frac{e ^{u}}{2} d u)

Take the constant out:

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

= 2 (- (- \frac{1}{2} \cdot \int_{- 25}^{0} e^{u} d u))

Use the common integral:

\int e^{u} d u = e^{u}

= 2 (- (- \frac{1}{2} [e^{u}]_{- 25}^{0}))

Simplify

2 (- (- \frac{1}{2} [e^{u}]_{- 25}^{0})) : [e^{u}]_{- 25}^{0}

= [e^{u}]_{- 25}^{0}

Compute the boundaries:

1 - \frac{1}{e ^{25}}

= 1 - \frac{1}{e ^{25}}

\int_{- 1}^{1} (x^{3} - 1) d x

\int_{1}^{5} (\frac{x}{5} + \frac{y}{20}) d y

\int_{- 3}^{6} 3 x - 5 d x

\int_{0}^{1} (7 x^{2} - 8 x + 6) d x

\int_{1}^{5} x^{2} - 2 x - 15 d x

What is the integral from 0 to 5 of 2te^{-t^2} ?
The integral from 0 to 5 of 2te^{-t^2} is 1-1/(e^{25)}