What is the integral from 0 to 2 of 2x^4e^{-x} ?

The integral from 0 to 2 of 2x^4e^{-x} is 2(-(168)/(e^2)+24)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to 2 of 2x^4e^{-x}

\int_{0}^{2} 2 x^{4} e^{- x} d x

2 (- \frac{168}{e ^{2}} + 24)

Decimal Notation

2.52734 \dots

Solution steps

\int_{0}^{2} 2 x^{4} e^{- x} d x

Take the constant out:

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

= 2 \cdot \int_{0}^{2} x^{4} e^{- x} d x

Apply Integration By Parts

: [- e^{- x} x^{4} - \int - 4 e^{- x} x^{3} d x]_{0}^{2}

= 2 [- e^{- x} x^{4} - \int - 4 e^{- x} x^{3} d x]_{0}^{2}

\int - 4 e^{- x} x^{3} d x = - 4 (- e^{- x} x^{3} - 3 (e^{- x} x^{2} - 2 (- e^{- x} x - e^{- x})))

= 2 [- e^{- x} x^{4} - (- 4 (- e^{- x} x^{3} - 3 (e^{- x} x^{2} - 2 (- e^{- x} x - e^{- x}))))]_{0}^{2}

Simplify

= 2 [- e^{- x} x^{4} + 4 (- e^{- x} x^{3} - 3 (e^{- x} x^{2} - 2 (- e^{- x} x - e^{- x})))]_{0}^{2}

Compute the boundaries:

- \frac{168}{e ^{2}} + 24

= 2 (- \frac{168}{e ^{2}} + 24)

y = \frac{1}{14 x ^{2}}, y = x, y = \frac{x}{10}

\frac{d}{d y} (x + y z)

x \cdot y^{^{'}} = 1 - y^{2}

\frac{\partial}{\partial y} (arctan (x y))

\frac{d}{d x} (\frac{a x + b y ( x )}{c x + d y})

What is the integral from 0 to 2 of 2x^4e^{-x} ?
The integral from 0 to 2 of 2x^4e^{-x} is 2(-(168)/(e^2)+24)