What is the integral from 1 to 25 of 6/(sqrt(x)) ?

The integral from 1 to 25 of 6/(sqrt(x)) is 48

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 1 to 25 of 6/(sqrt(x))

\int_{1}^{25} \frac{6}{x} d x

48

Solution steps

\int_{1}^{25} \frac{6}{x} d x

Take the constant out:

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

= 6 \cdot \int_{1}^{25} \frac{1}{x} d x

Apply the Power Rule

: [2 x^{\frac{1}{2}}]_{1}^{25}

= 6 [2 x^{\frac{1}{2}}]_{1}^{25}

Apply radical rule:

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

= 6 [2 x]_{1}^{25}

Compute the boundaries:

8

= 6 \cdot 8

Simplify

= 48

\int_{0}^{1} a x^{3} d x

\int_{0}^{\frac{π}{3}} 3 sin (θ) d θ

\int_{0}^{1} [x - x^{2}] d x

\int_{- 3}^{0} x^{2} - 20 d x

\int_{1 - 6 x}^{1} \frac{u ^{3}}{1 + u ^{2}} d u

What is the integral from 1 to 25 of 6/(sqrt(x)) ?
The integral from 1 to 25 of 6/(sqrt(x)) is 48