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

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

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

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

\frac{2}{3}

Decimal Notation

0.66666 \dots

Solution steps

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

Apply u-substitution

: \int_{0}^{1} u d u

= \int_{0}^{1} u d u

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{0}^{1}

= [\frac{2}{3} u^{\frac{3}{2}}]_{0}^{1}

Compute the boundaries:

\frac{2}{3}

= \frac{2}{3}

\frac{3 s}{( s ^{2} + 4 )}

y^{^{''''}} + 2 y^{^{''}} + y = 0

\frac{d}{d x} ((\frac{10}{4 + x})^{\frac{1}{2}})

- \frac{1}{4} x^{4} + 6 x^{2}

\frac{d}{d x} (\frac{x ^{4}}{12})

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