What is the integral from x to 0 of e^{-7t} ?

The integral from x to 0 of e^{-7t} is -1/7 (1-e^{-7x})

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral from x to 0 of e^{-7t}

\int_{x}^{0} e^{- 7 t} d t

- \frac{1}{7} (1 - e^{- 7 x})

Solution steps

\int_{x}^{0} e^{- 7 t} d t

Apply u-substitution

= \int_{- 7 x}^{0} - \frac{1}{7} e^{u} d u

Take the constant out:

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

= - \frac{1}{7} \cdot \int_{- 7 x}^{0} e^{u} d u

Use the common integral:

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

= - \frac{1}{7} [e^{u}]_{- 7 x}^{0}

Compute the boundaries:

1 - e^{- 7 x}

= - \frac{1}{7} (1 - e^{- 7 x})

What is the integral from x to 0 of e^{-7t} ?
The integral from x to 0 of e^{-7t} is -1/7 (1-e^{-7x})