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

The integral from 1 to 2 of sqrt(x)(x-1) is (4sqrt(2)+4)/(15)

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

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

\frac{4 2 + 4}{15}

Decimal Notation

0.64379 \dots

Solution steps

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

Expand

x (x - 1) : x^{\frac{3}{2}} - x

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

\int_{1}^{2} x^{\frac{3}{2}} d x = \frac{8 2 - 2}{5}

\int_{1}^{2} x d x = \frac{4 2 - 2}{3}

= \frac{8 2 - 2}{5} - \frac{4 2 - 2}{3}

Simplify

\frac{8 2 - 2}{5} - \frac{4 2 - 2}{3} : \frac{4 2 + 4}{15}

= \frac{4 2 + 4}{15}

\frac{\partial}{\partial y} (y e^{x})

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\int_{0}^{y} 8 d x

f (x) = (3 x - 5)^{\frac{1}{2}}, at x = 2

What is the integral from 1 to 2 of sqrt(x)(x-1) ?
The integral from 1 to 2 of sqrt(x)(x-1) is (4sqrt(2)+4)/(15)