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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral from 0 to 2 of 2^{-x}

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

\frac{3}{4 ln ( 2 )}

Decimal

1.08202 \dots

Solution steps

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

Apply u-substitution

= \int_{0}^{- 2} - 2^{u} d u

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

= - \int_{- 2}^{0} - 2^{u} d u

Take the constant out:

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

= - (- \int_{- 2}^{0} 2^{u} d u)

\int a^{x} d x = \frac{a ^{x}}{ln a}

= - (- [\frac{2 ^{u}}{ln ( 2 )}]_{- 2}^{0})

Simplify

= [\frac{2 ^{u}}{ln ( 2 )}]_{- 2}^{0}

Compute the boundaries:

\frac{3}{4 ln ( 2 )}

= \frac{3}{4 ln ( 2 )}

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