What is the integral from 0 to 9 of sqrt(1+9x) ?

The integral from 0 to 9 of sqrt(1+9x) is (164sqrt(82)-2)/(27)

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integral from 0 to 9 of sqrt(1+9x)

\int_{0}^{9} 1 + 9 x d x

\frac{164 82 - 2}{27}

Decimal

54.92900 \dots

Solution steps

\int_{0}^{9} 1 + 9 x d x

Apply u-substitution

: \int_{1}^{82} \frac{u}{9} d u

= \int_{1}^{82} \frac{u}{9} d u

Take the constant out:

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

= \frac{1}{9} \cdot \int_{1}^{82} u d u

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{82}

= \frac{1}{9} [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{82}

Compute the boundaries:

\frac{164 82}{3} - \frac{2}{3}

= \frac{1}{9} (\frac{164 82}{3} - \frac{2}{3})

Simplify

= \frac{164 82 - 2}{27}

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What is the integral from 0 to 9 of sqrt(1+9x) ?
The integral from 0 to 9 of sqrt(1+9x) is (164sqrt(82)-2)/(27)