What is the integral from 9 to 12 of (9-x)(x-12) ?

The integral from 9 to 12 of (9-x)(x-12) is 9/2

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integral from 9 to 12 of (9-x)(x-12)

\int_{9}^{12} (9 - x) (x - 12) d x

\frac{9}{2}

Decimal

4.5

Solution steps

\int_{9}^{12} (9 - x) (x - 12) d x

Expand

(9 - x) (x - 12) : 21 x - x^{2} - 108

= \int_{9}^{12} 21 x - x^{2} - 108 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_{9}^{12} 21 x d x - \int_{9}^{12} x^{2} d x - \int_{9}^{12} 108 d x

\int_{9}^{12} 21 x d x = \frac{1323}{2}

\int_{9}^{12} x^{2} d x = 333

\int_{9}^{12} 108 d x = 324

= \frac{1323}{2} - 333 - 324

Simplify

\frac{1323}{2} - 333 - 324 : \frac{9}{2}

= \frac{9}{2}

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What is the integral from 9 to 12 of (9-x)(x-12) ?
The integral from 9 to 12 of (9-x)(x-12) is 9/2