What is the integral from 0 to 1 of x^2e^{-7x} ?

The integral from 0 to 1 of x^2e^{-7x} is (2e^7-65)/(343e^7)

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 x^2e^{-7x}

\int_{0}^{1} x^{2} e^{- 7 x} d x

\frac{2 e ^{7} - 65}{343 e ^{7}}

Decimal Notation

0.00565 \dots

Solution steps

\int_{0}^{1} x^{2} e^{- 7 x} d x

Apply u-substitution

: \int_{0}^{- 7} - \frac{e ^{u} u ^{2}}{343} d u

= \int_{0}^{- 7} - \frac{e ^{u} u ^{2}}{343} d u

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

= - \int_{- 7}^{0} - \frac{e ^{u} u ^{2}}{343} d u

Take the constant out:

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

= - (- \frac{1}{343} \cdot \int_{- 7}^{0} e^{u} u^{2} d u)

Apply Integration By Parts

: [u^{2} e^{u} - \int 2 u e^{u} d u]_{- 7}^{0}

= - (- \frac{1}{343} [u^{2} e^{u} - \int 2 u e^{u} d u]_{- 7}^{0})

\int 2 u e^{u} d u = 2 (e^{u} u - e^{u})

= - (- \frac{1}{343} [u^{2} e^{u} - 2 (e^{u} u - e^{u})]_{- 7}^{0})

Simplify

= \frac{1}{343} [e^{u} u^{2} - 2 (e^{u} u - e^{u})]_{- 7}^{0}

Compute the boundaries:

2 - \frac{65}{e ^{7}}

= \frac{1}{343} (2 - \frac{65}{e ^{7}})

Simplify

= \frac{2 e ^{7} - 65}{343 e ^{7}}

\int_{1}^{2} k x^{2} d x

\int_{1}^{4} 3 x x + 1 d x

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

\int_{0}^{1} x^{8} e^{- x^{3}} d x

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

What is the integral from 0 to 1 of x^2e^{-7x} ?
The integral from 0 to 1 of x^2e^{-7x} is (2e^7-65)/(343e^7)