What is the integral from 4 to 9 of 1/(3sqrt(x)) ?

The integral from 4 to 9 of 1/(3sqrt(x)) is 2/3

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integral from 4 to 9 of 1/(3sqrt(x))

\int_{4}^{9} \frac{1}{3 x} d x

\frac{2}{3}

Decimal Notation

0.66666 \dots

Solution steps

\int_{4}^{9} \frac{1}{3 x} d x

Take the constant out:

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

= \frac{1}{3} \cdot \int_{4}^{9} \frac{1}{x} d x

Apply the Power Rule

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

= \frac{1}{3} [2 x^{\frac{1}{2}}]_{4}^{9}

Apply radical rule:

= \frac{1}{3} [2 x]_{4}^{9}

Compute the boundaries:

2

= \frac{1}{3} \cdot 2

Simplify

= \frac{2}{3}

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What is the integral from 4 to 9 of 1/(3sqrt(x)) ?
The integral from 4 to 9 of 1/(3sqrt(x)) is 2/3