What is the integral from 1 to 9 of 2/(sqrt(x)) ?

The integral from 1 to 9 of 2/(sqrt(x)) is 8

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} \frac{2}{x} d x

8

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \int_{1}^{9} \frac{1}{x} d x

Apply the Power Rule

: [2 x^{\frac{1}{2}}]_{1}^{9}

= 2 [2 x^{\frac{1}{2}}]_{1}^{9}

Apply radical rule:

a^{\frac{1}{n}} = n a

= 2 [2 x]_{1}^{9}

Compute the boundaries:

4

= 2 \cdot 4

Simplify

= 8

\frac{d}{d x} (\frac{2}{3} x^{2} (x^{3} - 1))

\int (x^{3} + x^{2} - x) d x

\int \frac{cos ( x )}{9 - sin ( x )} d x

\frac{d}{d x} (3 x (10 x^{3} - \frac{6}{x + 10}))

\frac{d}{d x} (- 5 cos (5 x))

What is the integral from 1 to 9 of 2/(sqrt(x)) ?
The integral from 1 to 9 of 2/(sqrt(x)) is 8