What is the integral from 0 to 2 of 8e^{-3x} ?

The integral from 0 to 2 of 8e^{-3x} is (8(e^6-1))/(3e^6)

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 8e^{-3x}

\int_{0}^{2} 8 e^{- 3 x} d x

\frac{8 ( e ^{6} - 1 )}{3 e ^{6}}

Decimal Notation

2.66005 \dots

Solution steps

\int_{0}^{2} 8 e^{- 3 x} d x

Take the constant out:

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

= 8 \cdot \int_{0}^{2} e^{- 3 x} d x

Apply u-substitution

: \int_{0}^{- 6} - \frac{1}{3} e^{u} d u

= 8 \cdot \int_{0}^{- 6} - \frac{1}{3} e^{u} d u

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

= 8 (- \int_{- 6}^{0} - \frac{1}{3} e^{u} d u)

Take the constant out:

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

= 8 (- (- \frac{1}{3} \cdot \int_{- 6}^{0} e^{u} d u))

Use the common integral:

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

= 8 (- (- \frac{1}{3} [e^{u}]_{- 6}^{0}))

Simplify

8 (- (- \frac{1}{3} [e^{u}]_{- 6}^{0})) : \frac{8}{3} [e^{u}]_{- 6}^{0}

= \frac{8}{3} [e^{u}]_{- 6}^{0}

Compute the boundaries:

1 - \frac{1}{e ^{6}}

= \frac{8}{3} (1 - \frac{1}{e ^{6}})

Simplify

= \frac{8 ( e ^{6} - 1 )}{3 e ^{6}}

x \to - 3 lim (9 - x^{2})

k = 0 \sum \infty 3 (\frac{x}{4})^{k}

(\frac{x ^{2} + 1}{x ^{2} - 1})^{3}

h \to 0 lim (3 h + 6)

x \to 0 lim (\frac{cos ( 4 x ^{2} )}{x})

What is the integral from 0 to 2 of 8e^{-3x} ?
The integral from 0 to 2 of 8e^{-3x} is (8(e^6-1))/(3e^6)