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

The integral from 1 to 2 of sqrt(x-1)-(x-1) is 1/6

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

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

\frac{1}{6}

Decimal Notation

0.16666 \dots

Solution steps

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

Simplify

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int_{1}^{2} x - 1 d x = \frac{2}{3}

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

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

= \frac{2}{3} - \frac{3}{2} + 1

Simplify

\frac{2}{3} - \frac{3}{2} + 1 : \frac{1}{6}

= \frac{1}{6}

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What is the integral from 1 to 2 of sqrt(x-1)-(x-1) ?
The integral from 1 to 2 of sqrt(x-1)-(x-1) is 1/6