What is the integral from 0 to 3 of (x-1)(x-2)(x-3) ?

The integral from 0 to 3 of (x-1)(x-2)(x-3) is -9/4

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integral from 0 to 3 of (x-1)(x-2)(x-3)

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

- \frac{9}{4}

Decimal Notation

- 2.25

Solution steps

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

Expand

(x - 1) (x - 2) (x - 3) : x^{3} - 6 x^{2} + 11 x - 6

= \int_{0}^{3} x^{3} - 6 x^{2} + 11 x - 6 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_{0}^{3} x^{3} d x - \int_{0}^{3} 6 x^{2} d x + \int_{0}^{3} 11 x d x - \int_{0}^{3} 6 d x

\int_{0}^{3} x^{3} d x = \frac{81}{4}

\int_{0}^{3} 6 x^{2} d x = 54

\int_{0}^{3} 11 x d x = \frac{99}{2}

\int_{0}^{3} 6 d x = 18

= \frac{81}{4} - 54 + \frac{99}{2} - 18

Simplify

\frac{81}{4} - 54 + \frac{99}{2} - 18 : - \frac{9}{4}

= - \frac{9}{4}

(- 7, - 9), (0, - 7)

\int (4 - y^{2})^{\frac{1}{2}} d y

\frac{d}{d x} (sin (1 - x) cos^{3} (x))

f (x) = \frac{e ^{x}}{2 + 5 x}

x + \frac{y d y}{d x} = 0

What is the integral from 0 to 3 of (x-1)(x-2)(x-3) ?
The integral from 0 to 3 of (x-1)(x-2)(x-3) is -9/4