What is the integral from 0 to 21 of-x(x-21)(x+1) ?

The integral from 0 to 21 of-x(x-21)(x+1) is 71001/4

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integral from 0 to 21 of-x(x-21)(x+1)

\int_{0}^{21} - x (x - 21) (x + 1) d x

\frac{71001}{4}

Decimal Notation

17750.25

Solution steps

\int_{0}^{21} - x (x - 21) (x + 1) d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - \int_{0}^{21} x (x - 21) (x + 1) d x

Expand

x (x - 21) (x + 1) : x^{3} - 20 x^{2} - 21 x

= - \int_{0}^{21} x^{3} - 20 x^{2} - 21 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_{0}^{21} x^{3} d x - \int_{0}^{21} 20 x^{2} d x - \int_{0}^{21} 21 x d x)

\int_{0}^{21} x^{3} d x = \frac{194481}{4}

\int_{0}^{21} 20 x^{2} d x = 61740

\int_{0}^{21} 21 x d x = \frac{9261}{2}

= - (\frac{194481}{4} - 61740 - \frac{9261}{2})

Simplify

- (\frac{194481}{4} - 61740 - \frac{9261}{2}) : \frac{71001}{4}

= \frac{71001}{4}

\frac{d}{d x} ((x)^{x})

y^{^{'}} + \frac{y}{10} = 500

\frac{d N}{d t} = N (1 - 0.0002 N)

\frac{d}{d x} (c_{1} cos (2 x) + c_{2} sin (2 x))

\frac{d}{d x} (sec (\frac{1}{2} x) tan (\frac{1}{2} x))

What is the integral from 0 to 21 of-x(x-21)(x+1) ?
The integral from 0 to 21 of-x(x-21)(x+1) is 71001/4