What is the integral from 0 to x of λe^{-λx} ?

The integral from 0 to x of λe^{-λx} is -e^{-λx}+1

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to x of λe^{-λx}

\int_{0}^{x} λ e^{- λ x} d x

- e^{- λ x} + 1

Solution steps

\int_{0}^{x} λ e^{- λ x} d x

Take the constant out:

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

= λ \cdot \int_{0}^{x} e^{- λ x} d x

Apply u-substitution

: \int_{0}^{- λ x} - \frac{e ^{u}}{λ} d u

= λ \cdot \int_{0}^{- λ x} - \frac{e ^{u}}{λ} d u

Take the constant out:

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

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

Use the common integral:

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

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

Simplify

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

= - [e^{u}]_{0}^{- λ x}

Compute the boundaries:

e^{- λ x} - 1

= - (e^{- λ x} - 1)

Simplify

= - e^{- λ x} + 1

\int_{0}^{8} \frac{1}{1 + 1 + x} d x

\int_{0}^{1} x^{3} (2 + x^{4})^{3} d x

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

\int_{0}^{1} 30 x^{4} (1 - x)^{2} d x

\int_{3}^{s i n (x + y)} e^{t^{3}} d t

What is the integral from 0 to x of λe^{-λx} ?
The integral from 0 to x of λe^{-λx} is -e^{-λx}+1