What is the integral from 0 to 7 of t*e^{-2t} ?

The integral from 0 to 7 of t*e^{-2t} is -(-e^{14}+15)/(4e^{14)}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to 7 of t*e^{-2t}

\int_{0}^{7} t \cdot e^{- 2 t} d t

- \frac{- e ^{14} + 15}{4 e ^{14}}

Decimal Notation

0.24999 \dots

Solution steps

\int_{0}^{7} t e^{- 2 t} d t

Apply u-substitution

: \int_{0}^{- 14} \frac{e ^{u} u}{4} d u

= \int_{0}^{- 14} \frac{e ^{u} u}{4} d u

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

= - \int_{- 14}^{0} \frac{e ^{u} u}{4} d u

Take the constant out:

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

= - \frac{1}{4} \cdot \int_{- 14}^{0} e^{u} u d u

Apply Integration By Parts

: [e^{u} u - \int e^{u} d u]_{- 14}^{0}

= - \frac{1}{4} [e^{u} u - \int e^{u} d u]_{- 14}^{0}

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

= - \frac{1}{4} [e^{u} u - e^{u}]_{- 14}^{0}

Compute the boundaries:

- 1 + \frac{15}{e ^{14}}

= - \frac{1}{4} (- 1 + \frac{15}{e ^{14}})

Simplify

= - \frac{- e ^{14} + 15}{4 e ^{14}}

\int_{2}^{\infty} I n (2 x - 3) d x

\int_{0}^{5} \frac{1}{( x - 1 ) ^{3}} d x

\int_{- 1}^{2} w^{2} e^{4 w} d w

\int_{3}^{6} \frac{1}{( 3 x - 12 ) ^{3}} d x

\int_{x^{2} + 2}^{3 x} 1 d y

What is the integral from 0 to 7 of t*e^{-2t} ?
The integral from 0 to 7 of t*e^{-2t} is -(-e^{14}+15)/(4e^{14)}