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

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

Popular >

integral from 1 to 2 of xsqrt((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}

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

\int_{\frac{π}{2}}^{\frac{3 π}{2}} cos (x) d x

\int_{0}^{a} (3 x^{2} - x + 1) d x

\int_{0}^{0.3} 3 e^{- x^{2}} d x

\int_{0}^{2} x 2 x^{2} + 1 d x

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