What is the integral from 0 to 1 of (x+8)(x-9) ?

The integral from 0 to 1 of (x+8)(x-9) is -433/6

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integral from 0 to 1 of (x+8)(x-9)

\int_{0}^{1} (x + 8) (x - 9) d x

- \frac{433}{6}

Decimal

- 72.16666 \dots

Solution steps

\int_{0}^{1} (x + 8) (x - 9) d x

Expand

(x + 8) (x - 9) : x^{2} - x - 72

= \int_{0}^{1} x^{2} - x - 72 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}^{1} x^{2} d x - \int_{0}^{1} x d x - \int_{0}^{1} 72 d x

\int_{0}^{1} x^{2} d x = \frac{1}{3}

\int_{0}^{1} x d x = \frac{1}{2}

\int_{0}^{1} 72 d x = 72

= \frac{1}{3} - \frac{1}{2} - 72

Simplify

\frac{1}{3} - \frac{1}{2} - 72 : - \frac{433}{6}

= - \frac{433}{6}

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What is the integral from 0 to 1 of (x+8)(x-9) ?
The integral from 0 to 1 of (x+8)(x-9) is -433/6