What is the integral from 0 to infinity of 1/(3^x) ?

The integral from 0 to infinity of 1/(3^x) is 1/(ln(3))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to infinity of 1/(3^x)

\int_{0}^{\infty} \frac{1}{3 ^{x}} d x

\frac{1}{ln ( 3 )}

Decimal

0.91023 \dots

Solution steps

\int_{0}^{\infty} \frac{1}{3 ^{x}} d x

Compute the indefinite integral:

\int \frac{1}{3 ^{x}} d x = - \frac{1}{ln ( 3 ) \cdot 3 ^{x}} + C

Compute the boundaries:

\int_{0}^{\infty} \frac{1}{3 ^{x}} d x = 0 - (- \frac{1}{ln ( 3 )})

= 0 - (- \frac{1}{ln ( 3 )})

Simplify

= \frac{1}{ln ( 3 )}

\int_{0}^{6} (6) d x

\int_{- 3}^{3} sinh (x) + cosh (x) d x

\int_{- 2}^{1} 2 x^{2} + 2 x - 4 d x

\int_{1}^{3} r^{2} d r

\int_{- 2}^{2} ∣ x ∣ - (6 - x^{2}) d x

What is the integral from 0 to infinity of 1/(3^x) ?
The integral from 0 to infinity of 1/(3^x) is 1/(ln(3))