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

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

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

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

\frac{3 3 - 1}{3}

Decimal

1.39871 \dots

Solution steps

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

Apply u-substitution

: \int_{1}^{3} \frac{u}{2} d u

= \int_{1}^{3} \frac{u}{2} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int_{1}^{3} u d u

Apply the Power Rule

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

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

Compute the boundaries:

23 - \frac{2}{3}

= \frac{1}{2} (23 - \frac{2}{3})

Simplify

= \frac{3 3 - 1}{3}

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