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

The integral from 0 to 2 of 3e^{-2x} is (3(e^4-1))/(2e^4)

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 3e^{-2x}

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

\frac{3 ( e ^{4} - 1 )}{2 e ^{4}}

Decimal Notation

1.47252 \dots

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

: \int_{0}^{- 4} - \frac{1}{2} e^{u} d u

= 3 \cdot \int_{0}^{- 4} - \frac{1}{2} e^{u} d u

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

= 3 (- \int_{- 4}^{0} - \frac{1}{2} e^{u} d u)

Take the constant out:

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

= 3 (- (- \frac{1}{2} \cdot \int_{- 4}^{0} e^{u} d u))

Use the common integral:

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

= 3 (- (- \frac{1}{2} [e^{u}]_{- 4}^{0}))

Simplify

3 (- (- \frac{1}{2} [e^{u}]_{- 4}^{0})) : \frac{3}{2} [e^{u}]_{- 4}^{0}

= \frac{3}{2} [e^{u}]_{- 4}^{0}

Compute the boundaries:

1 - \frac{1}{e ^{4}}

= \frac{3}{2} (1 - \frac{1}{e ^{4}})

Simplify

= \frac{3 ( e ^{4} - 1 )}{2 e ^{4}}

\frac{d}{d x} (x^{5} - 5 x^{3} + 4 x)

e^{8 x^{2}}

\frac{d}{d x} ((2 x - x^{2})^{2} + 9)

\int (x - 2) ln (x) d x

4 x^{2} + y^{2} = 25, (2, 3)

What is the integral from 0 to 2 of 3e^{-2x} ?
The integral from 0 to 2 of 3e^{-2x} is (3(e^4-1))/(2e^4)