What is the integral from 4 to 9 of 1/(sqrt(x)) ?

The integral from 4 to 9 of 1/(sqrt(x)) is 2

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integral from 4 to 9 of 1/(sqrt(x))

\int_{4}^{9} \frac{1}{x} d x

2

Solution steps

\int_{4}^{9} \frac{1}{x} d x

Apply the Power Rule

: [2 x^{\frac{1}{2}}]_{4}^{9}

= [2 x^{\frac{1}{2}}]_{4}^{9}

Apply radical rule:

= [2 x]_{4}^{9}

Compute the boundaries:

2

= 2

x \to 5 lim (\frac{x - 5}{∣ x - 5 ∣})

\int_{0}^{2} 1 + e^{- 2 y} d y

\int \frac{9}{1 - x ^{2}} d x

x \to 15 lim (2)

\frac{d}{d x} (\frac{x ^{3} + 7}{x})

What is the integral from 4 to 9 of 1/(sqrt(x)) ?
The integral from 4 to 9 of 1/(sqrt(x)) is 2