What is the integral from 1 to 9 of (1-x)(x-9) ?

The integral from 1 to 9 of (1-x)(x-9) is 256/3

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integral from 1 to 9 of (1-x)(x-9)

\int_{1}^{9} (1 - x) (x - 9) d x

\frac{256}{3}

Decimal

85.33333 \dots

Solution steps

\int_{1}^{9} (1 - x) (x - 9) d x

Expand

(1 - x) (x - 9) : 10 x - x^{2} - 9

= \int_{1}^{9} 10 x - x^{2} - 9 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}^{9} 10 x d x - \int_{1}^{9} x^{2} d x - \int_{1}^{9} 9 d x

\int_{1}^{9} 10 x d x = 400

\int_{1}^{9} x^{2} d x = \frac{728}{3}

\int_{1}^{9} 9 d x = 72

= 400 - \frac{728}{3} - 72

Simplify

400 - \frac{728}{3} - 72 : \frac{256}{3}

= \frac{256}{3}

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What is the integral from 1 to 9 of (1-x)(x-9) ?
The integral from 1 to 9 of (1-x)(x-9) is 256/3