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

The integral from 0 to x of 2e^{-2t} is -e^{-2x}+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 2e^{-2t}

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

- e^{- 2 x} + 1

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

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

= 2 \cdot \int_{0}^{- 2 x} - \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

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

Use the common integral:

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

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

Simplify

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

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

Compute the boundaries:

e^{- 2 x} - 1

= - (e^{- 2 x} - 1)

Simplify

= - e^{- 2 x} + 1

\int_{0}^{5} 6 x e^{3 x^{2}} d x

\int_{8}^{12} (8 - x) (x - 12) d x

\int_{0}^{π} tan (\frac{θ}{4}) d θ

\int_{5}^{6} (\frac{1}{x ^{2}} - 1) d x

\int_{0}^{1} x (2 - x^{2})^{5} d x

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