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

The integral from 1 to 2 of x(sqrt(x-1)) is 16/15

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

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

\frac{16}{15}

Decimal Notation

1.06666 \dots

Solution steps

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

Apply u-substitution

: \int_{0}^{1} (u + 1) u d u

= \int_{0}^{1} (u + 1) u d u

Expand

(u + 1) u : u^{\frac{3}{2}} + u

= \int_{0}^{1} u^{\frac{3}{2}} + u d u

Apply the Sum Rule:

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

= \int_{0}^{1} u^{\frac{3}{2}} d u + \int_{0}^{1} u d u

\int_{0}^{1} u^{\frac{3}{2}} d u = \frac{2}{5}

\int_{0}^{1} u d u = \frac{2}{3}

= \frac{2}{5} + \frac{2}{3}

Simplify

\frac{2}{5} + \frac{2}{3} : \frac{16}{15}

= \frac{16}{15}

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What is the integral from 1 to 2 of x(sqrt(x-1)) ?
The integral from 1 to 2 of x(sqrt(x-1)) is 16/15