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

The integral from 1 to 16 of 8/(sqrt(x)) is 48

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

\int_{1}^{16} \frac{8}{x} d x

48

Solution steps

\int_{1}^{16} \frac{8}{x} d x

Take the constant out:

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

= 8 \cdot \int_{1}^{16} \frac{1}{x} d x

Apply the Power Rule

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

= 8 [2 x^{\frac{1}{2}}]_{1}^{16}

Apply radical rule:

= 8 [2 x]_{1}^{16}

Compute the boundaries:

6

= 8 \cdot 6

Simplify

= 48

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

f (x) = e^{- 4 x}

\frac{d}{d x} (\frac{14 x + 8 x ^{2}}{( 7 + 8 x ) ^{2}})

x \to - 2 lim (\frac{x}{x ^{2} + 4})

\int e^{x^{2}} \frac{x}{3} d x

What is the integral from 1 to 16 of 8/(sqrt(x)) ?
The integral from 1 to 16 of 8/(sqrt(x)) is 48