What is the integral from 1 to 9 of 6/(sqrt(x)) ?

The integral from 1 to 9 of 6/(sqrt(x)) is 24

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integral from 1 to 9 of 6/(sqrt(x))

\int_{1}^{9} \frac{6}{x} d x

24

Solution steps

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

Apply the Power Rule

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

= 6 [2 x^{\frac{1}{2}}]_{1}^{9}

Apply radical rule:

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

= 6 [2 x]_{1}^{9}

Compute the boundaries:

4

= 6 \cdot 4

Simplify

= 24

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What is the integral from 1 to 9 of 6/(sqrt(x)) ?
The integral from 1 to 9 of 6/(sqrt(x)) is 24