What is the integral from 9 to 36 of 1/(sqrt(x)) ?

The integral from 9 to 36 of 1/(sqrt(x)) is 6

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integral from 9 to 36 of 1/(sqrt(x))

\int_{9}^{36} \frac{1}{x} d x

6

Solution steps

\int_{9}^{36} \frac{1}{x} d x

Apply the Power Rule

: [2 x^{\frac{1}{2}}]_{9}^{36}

= [2 x^{\frac{1}{2}}]_{9}^{36}

Apply radical rule:

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

= [2 x]_{9}^{36}

Compute the boundaries:

6

= 6

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What is the integral from 9 to 36 of 1/(sqrt(x)) ?
The integral from 9 to 36 of 1/(sqrt(x)) is 6