What is the integral from 1 to 9 of 2^{sqrt(x)} ?

The integral from 1 to 9 of 2^{sqrt(x)} is (44ln(2)-12)/(ln^2(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 9 of 2^{sqrt(x)}

\int_{1}^{9} 2^{x} d x

\frac{44 ln ( 2 ) - 12}{ln ^{2} ( 2 )}

Decimal Notation

38.50215 \dots

Solution steps

\int_{1}^{9} 2^{x} d x

Apply u-substitution

: \int_{1}^{3} 2^{u + 1} u d u

= \int_{1}^{3} 2^{u + 1} u d u

Apply Integration By Parts

: [\frac{2 ^{u + 1} u}{ln ( 2 )} - \int \frac{2 ^{u + 1}}{ln ( 2 )} d u]_{1}^{3}

= [\frac{2 ^{u + 1} u}{ln ( 2 )} - \int \frac{2 ^{u + 1}}{ln ( 2 )} d u]_{1}^{3}

\int \frac{2 ^{u + 1}}{ln ( 2 )} d u = \frac{1}{ln ^{2} ( 2 )} \cdot 2^{1 + u}

= [\frac{2 ^{u + 1} u}{ln ( 2 )} - \frac{1}{ln ^{2} ( 2 )} \cdot 2^{1 + u}]_{1}^{3}

Compute the boundaries:

\frac{44 ln ( 2 ) - 12}{ln ^{2} ( 2 )}

= \frac{44 ln ( 2 ) - 12}{ln ^{2} ( 2 )}

\int \frac{2}{( x + 2 ) ^{2}} d x

\frac{d}{d x} (3 x^{4} + 4 x^{3})

\frac{\partial}{\partial x} (60 - x^{2} y^{3})

\int_{- 125}^{27} \frac{1}{x ^{\frac{1}{3}}} d x

(2 x^{7} - 5 x^{4} + 2)^{25}

What is the integral from 1 to 9 of 2^{sqrt(x)} ?
The integral from 1 to 9 of 2^{sqrt(x)} is (44ln(2)-12)/(ln^2(2))