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

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

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

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

\frac{1}{4}

Solution steps

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

Expand

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

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

\int_{1}^{2} 3 x^{2} d x = 7

\int_{1}^{2} x^{3} d x = \frac{15}{4}

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

= 7 - \frac{15}{4} - 3

Simplify

7 - \frac{15}{4} - 3 : \frac{1}{4}

= \frac{1}{4}

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