What is the integral from 4 to 16 of 1/(sqrt(z)) ?

The integral from 4 to 16 of 1/(sqrt(z)) is 4

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integral from 4 to 16 of 1/(sqrt(z))

\int_{4}^{16} \frac{1}{z} d z

4

Solution steps

\int_{4}^{16} \frac{1}{z} d z

Apply the Power Rule

: [2 z^{\frac{1}{2}}]_{4}^{16}

= [2 z^{\frac{1}{2}}]_{4}^{16}

Apply radical rule:

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

= [2 z]_{4}^{16}

Compute the boundaries:

4

= 4

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

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

\int_{- 3}^{0} (6 x^{3} - 3 x^{2} - 9 x - 6) d x

\int_{0}^{1} e^{- (2 x - 1)^{2}} d x

\int_{0}^{1} (x + x^{3}) d x

What is the integral from 4 to 16 of 1/(sqrt(z)) ?
The integral from 4 to 16 of 1/(sqrt(z)) is 4