What is the integral from-1 to 2 of 2^x ?

The integral from-1 to 2 of 2^x is 7/(2ln(2))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from-1 to 2 of 2^x

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

\frac{7}{2 ln ( 2 )}

Decimal

5.04943 \dots

Solution steps

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

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

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

Compute the boundaries:

\frac{7}{2 ln ( 2 )}

= \frac{7}{2 ln ( 2 )}

n = 0 \sum \infty e^{- \frac{n}{2}}

x + y^{2} = 4, x + y = 2

\frac{d}{d x} (uv w z (u, v, w, x))

x \to 3 - lim (x - 1)

x \to - 1 lim (6 x + 1)

What is the integral from-1 to 2 of 2^x ?
The integral from-1 to 2 of 2^x is 7/(2ln(2))