What is the integral from 0 to 16 of 1/(sqrt(x)) ?

The integral from 0 to 16 of 1/(sqrt(x)) is 8

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integral from 0 to 16 of 1/(sqrt(x))

\int_{0}^{16} \frac{1}{x} d x

8

Solution steps

\int_{0}^{16} \frac{1}{x} d x

Apply the Power Rule

: [2 x^{\frac{1}{2}}]_{0}^{16}

= [2 x^{\frac{1}{2}}]_{0}^{16}

Apply radical rule:

= [2 x]_{0}^{16}

Compute the boundaries:

8

= 8

y^{^{'}} = ry

y^{^{''}} = - \frac{2}{t ^{2}}

y = x^{2} + 5 x + 9, at x = 4

\frac{\partial}{\partial x} (yz e^{x z} + e^{x z} + x y e^{x z})

y = 1.2 (x + 0.1)^{0.5} + 4, 1, - 0.1

What is the integral from 0 to 16 of 1/(sqrt(x)) ?
The integral from 0 to 16 of 1/(sqrt(x)) is 8