What is the integral from 0 to 1 of sqrt(1+4x) ?

The integral from 0 to 1 of sqrt(1+4x) is (5sqrt(5)-1)/6

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integral from 0 to 1 of sqrt(1+4x)

\int_{0}^{1} 1 + 4 x d x

\frac{5 5 - 1}{6}

Decimal Notation

1.69672 \dots

Solution steps

\int_{0}^{1} 1 + 4 x d x

Apply u-substitution

: \int_{1}^{5} \frac{u}{4} d u

= \int_{1}^{5} \frac{u}{4} d u

Take the constant out:

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

= \frac{1}{4} \cdot \int_{1}^{5} u d u

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{5}

= \frac{1}{4} [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{5}

Compute the boundaries:

\frac{10 5}{3} - \frac{2}{3}

= \frac{1}{4} (\frac{10 5}{3} - \frac{2}{3})

Simplify

= \frac{5 5 - 1}{6}

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What is the integral from 0 to 1 of sqrt(1+4x) ?
The integral from 0 to 1 of sqrt(1+4x) is (5sqrt(5)-1)/6