What is the integral from 6 to 9 of (6-x)(x-9) ?

The integral from 6 to 9 of (6-x)(x-9) is 9/2

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integral from 6 to 9 of (6-x)(x-9)

\int_{6}^{9} (6 - x) (x - 9) d x

\frac{9}{2}

Decimal

4.5

Solution steps

\int_{6}^{9} (6 - x) (x - 9) d x

Expand

(6 - x) (x - 9) : 15 x - x^{2} - 54

= \int_{6}^{9} 15 x - x^{2} - 54 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_{6}^{9} 15 x d x - \int_{6}^{9} x^{2} d x - \int_{6}^{9} 54 d x

\int_{6}^{9} 15 x d x = \frac{675}{2}

\int_{6}^{9} x^{2} d x = 171

\int_{6}^{9} 54 d x = 162

= \frac{675}{2} - 171 - 162

Simplify

\frac{675}{2} - 171 - 162 : \frac{9}{2}

= \frac{9}{2}

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What is the integral from 6 to 9 of (6-x)(x-9) ?
The integral from 6 to 9 of (6-x)(x-9) is 9/2