What is the integral from 0 to 1 of 4e^{-2x} ?

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

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 4e^{-2x}

\int_{0}^{1} 4 e^{- 2 x} d x

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

Solution steps

\int_{0}^{1} 4 e^{- 2 x} d x

Take the constant out:

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

= 4 \cdot \int_{0}^{1} e^{- 2 x} d x

Apply u-substitution

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

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

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

= 4 (- \int_{- 2}^{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

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

Use the common integral:

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

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

Simplify

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

= 2 [e^{u}]_{- 2}^{0}

Compute the boundaries:

1 - \frac{1}{e ^{2}}

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

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

\int_{0}^{x} sinh^{2} (t) d t

\int_{0}^{t} e^{- a x} d x

\int_{- 2}^{2} x (x^{2} + 1)^{2} d x

\int_{0}^{1} x^{2} cos (x) d x

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