What is the integral from 0 to 9 of 1/(sqrt(x)) ?

The integral from 0 to 9 of 1/(sqrt(x)) is 6

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integral from 0 to 9 of 1/(sqrt(x))

\int_{0}^{9} \frac{1}{x} d x

6

Solution steps

\int_{0}^{9} \frac{1}{x} d x

Apply the Power Rule

= [2 x^{\frac{1}{2}}]_{0}^{9}

Apply radical rule:

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

= [2 x]_{0}^{9}

Compute the boundaries:

6

= 6

y = sec (x) tan (x)

\int 7 e^{3 x} d x

\frac{\partial}{\partial y} (\frac{2}{2 x + 3 y})

f (x) = x^{2} sin (x)

\frac{d}{d x} (- x^{3} + 9 x^{2} - 6 x + 9)

What is the integral from 0 to 9 of 1/(sqrt(x)) ?
The integral from 0 to 9 of 1/(sqrt(x)) is 6