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

The integral from 1 to 16 of 1/(sqrt(x)) is 6

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

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

6

Solution steps

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

Apply the Power Rule

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

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

Apply radical rule:

= [2 x]_{1}^{16}

Compute the boundaries:

6

= 6

\int \frac{2 x}{3 ( x ^{2} + 1 )} d x

\frac{d}{d x} (x^{12} ln (x))

\int \frac{16}{x ( x ^{2} + 4 ) ^{2}} d x

\frac{d}{d x} ((x sin (2 x))^{\frac{3}{2}})

\int \frac{( x + 16 )}{( x ^{2} + 2 x - 8 )} d x

What is the integral from 1 to 16 of 1/(sqrt(x)) ?
The integral from 1 to 16 of 1/(sqrt(x)) is 6