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

The integral from 1 to 16 of 6/(sqrt(x)) is 36

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

\int_{1}^{16} \frac{6}{x} d x

36

Solution steps

\int_{1}^{16} \frac{6}{x} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 6 \cdot \int_{1}^{16} \frac{1}{x} d x

Apply the Power Rule

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

= 6 [2 x^{\frac{1}{2}}]_{1}^{16}

Apply radical rule:

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

= 6 [2 x]_{1}^{16}

Compute the boundaries:

6

= 6 \cdot 6

Simplify

= 36

\int_{2}^{5} \frac{1}{2 - 5 t} d t

9 y^{^{''}} - 6 y^{^{'}} + 4 y = 0

\frac{d}{d x} ((1 - x) e^{- x})

x \to 0 lim (\frac{4 sin ( 3 x )}{2 tan ( 5 x )})

x \to (\infty) lim (\frac{ln ( x )}{x})

What is the integral from 1 to 16 of 6/(sqrt(x)) ?
The integral from 1 to 16 of 6/(sqrt(x)) is 36