What is the integral from 1 to 16 of 1/(2sqrt(v)) ?

The integral from 1 to 16 of 1/(2sqrt(v)) is 3

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integral from 1 to 16 of 1/(2sqrt(v))

\int_{1}^{16} \frac{1}{2 v} d v

3

Solution steps

\int_{1}^{16} \frac{1}{2 v} d v

Take the constant out:

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

= \frac{1}{2} \cdot \int_{1}^{16} \frac{1}{v} d v

Apply the Power Rule

: [2 v^{\frac{1}{2}}]_{1}^{16}

= \frac{1}{2} [2 v^{\frac{1}{2}}]_{1}^{16}

Apply radical rule:

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

= \frac{1}{2} [2 v]_{1}^{16}

Compute the boundaries:

6

= \frac{1}{2} \cdot 6

Simplify

= 3

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What is the integral from 1 to 16 of 1/(2sqrt(v)) ?
The integral from 1 to 16 of 1/(2sqrt(v)) is 3