What is the integral from 0 to log_{2 of (6)}2^x ?

The integral from 0 to log_{2 of (6)}2^x is 5/(ln(2))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to log_{2 of (6)}2^x

\int_{0}^{l o g_{2} (6)} 2^{x} d x

\frac{5}{ln ( 2 )}

Decimal

7.21347 \dots

Solution steps

\int_{0}^{l o g_{2} (6)} 2^{x} d x

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

= [\frac{2 ^{x}}{ln ( 2 )}]_{0}^{l o g_{2} (6)}

Compute the boundaries:

\frac{5}{ln ( 2 )}

= \frac{5}{ln ( 2 )}

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

\int_{0}^{2 π} - 16 sin^{2} (θ) d θ

\int_{0}^{1} 2 y e^{- y^{2}} d y

\int_{0}^{2} x + y d y

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

What is the integral from 0 to log_{2 of (6)}2^x ?
The integral from 0 to log_{2 of (6)}2^x is 5/(ln(2))