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

The integral from 1 to 8 of sqrt(1+x) is 18-(4sqrt(2))/3

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

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

18 - \frac{4 2}{3}

Decimal

16.11438 \dots

Solution steps

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

Apply u-substitution

: \int_{2}^{9} u d u

= \int_{2}^{9} u d u

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{2}^{9}

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

Compute the boundaries:

18 - \frac{4 2}{3}

= 18 - \frac{4 2}{3}

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What is the integral from 1 to 8 of sqrt(1+x) ?
The integral from 1 to 8 of sqrt(1+x) is 18-(4sqrt(2))/3