What is the integral from 2 to 9 of sqrt(x-1) ?

The integral from 2 to 9 of sqrt(x-1) is (32sqrt(2)-2)/3

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integral from 2 to 9 of sqrt(x-1)

\int_{2}^{9} x - 1 d x

\frac{32 2 - 2}{3}

Decimal

14.41827 \dots

Solution steps

\int_{2}^{9} x - 1 d x

Apply u-substitution

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

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

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{8}

= [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{8}

Compute the boundaries:

\frac{32 2}{3} - \frac{2}{3}

= \frac{32 2}{3} - \frac{2}{3}

Simplify

= \frac{32 2 - 2}{3}

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What is the integral from 2 to 9 of sqrt(x-1) ?
The integral from 2 to 9 of sqrt(x-1) is (32sqrt(2)-2)/3