What is the integral from 0 to 1 of 1/(sqrt(1+8x)) ?

The integral from 0 to 1 of 1/(sqrt(1+8x)) is 1/2

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integral from 0 to 1 of 1/(sqrt(1+8x))

\int_{0}^{1} \frac{1}{1 + 8 x} d x

\frac{1}{2}

Decimal Notation

0.5

Solution steps

\int_{0}^{1} \frac{1}{1 + 8 x} d x

Apply u-substitution

: \int_{1}^{9} \frac{1}{8 u} d u

= \int_{1}^{9} \frac{1}{8 u} d u

Take the constant out:

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

= \frac{1}{8} \cdot \int_{1}^{9} \frac{1}{u} d u

Apply the Power Rule

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

= \frac{1}{8} [2 u^{\frac{1}{2}}]_{1}^{9}

Apply radical rule:

= \frac{1}{8} [2 u]_{1}^{9}

Compute the boundaries:

4

= \frac{1}{8} \cdot 4

Simplify

= \frac{1}{2}

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What is the integral from 0 to 1 of 1/(sqrt(1+8x)) ?
The integral from 0 to 1 of 1/(sqrt(1+8x)) is 1/2