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

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

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

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

\frac{32 2}{3}

Decimal Notation

15.08494 \dots

Solution steps

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

Apply u-substitution

: \int_{0}^{8} u d u

= \int_{0}^{8} u d u

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{0}^{8}

= [\frac{2}{3} u^{\frac{3}{2}}]_{0}^{8}

Compute the boundaries:

\frac{32 2}{3}

= \frac{32 2}{3}

\frac{d}{d x} (8 x + 3)

x \to 0 + lim (x^{t a n (x^{2})})

\frac{\partial}{\partial x} (2 k x^{2} y^{2} - x sin (y))

\frac{2 e ^{- 2 s}}{s ^{2} - 4}

e^{t} - t

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