What is the integral from 0 to x of (x^{-1/2})/6 ?

The integral from 0 to x of (x^{-1/2})/6 is (x^{1/2})/3

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integral from 0 to x of (x^{-1/2})/6

\int_{0}^{x} \frac{x ^{- \frac{1}{2}}}{6} d x

\frac{x ^{\frac{1}{2}}}{3}

Solution steps

\int_{0}^{x} \frac{x ^{- \frac{1}{2}}}{6} d x

Take the constant out:

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

= \frac{1}{6} \cdot \int_{0}^{x} x^{- \frac{1}{2}} d x

Apply the Power Rule

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

= \frac{1}{6} [2 x^{\frac{1}{2}}]_{0}^{x}

Compute the boundaries:

2 x^{\frac{1}{2}}

= \frac{1}{6} \cdot 2 x^{\frac{1}{2}}

Simplify

= \frac{x ^{\frac{1}{2}}}{3}

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What is the integral from 0 to x of (x^{-1/2})/6 ?
The integral from 0 to x of (x^{-1/2})/6 is (x^{1/2})/3