What is the integral from 0 to 1 of 2^{4x} ?

The integral from 0 to 1 of 2^{4x} is (15)/(4ln(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 1 of 2^{4x}

\int_{0}^{1} 2^{4 x} d x

\frac{15}{4 ln ( 2 )}

Decimal Notation

5.41010 \dots

Solution steps

\int_{0}^{1} 2^{4 x} d x

Simplify

2^{4 x} : 1 6^{x}

= \int_{0}^{1} 1 6^{x} d x

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

= [\frac{1 6 ^{x}}{ln ( 16 )}]_{0}^{1}

\frac{1 6 ^{x}}{ln ( 16 )} = \frac{2 ^{4 x - 2}}{ln ( 2 )}

= [\frac{2 ^{4 x - 2}}{ln ( 2 )}]_{0}^{1}

Compute the boundaries:

\frac{15}{4 ln ( 2 )}

= \frac{15}{4 ln ( 2 )}

\int_{- 2}^{2} 5 d x

\int_{3}^{7} x d x

\int_{1}^{2} x cos (\frac{nπ x}{2}) d x

\int_{0}^{\frac{π}{6}} (cos (3 θ))^{2} d θ

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

What is the integral from 0 to 1 of 2^{4x} ?
The integral from 0 to 1 of 2^{4x} is (15)/(4ln(2))