What is the integral from 0 to 1 of (x+2)(x-3) ?

The integral from 0 to 1 of (x+2)(x-3) is -37/6

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integral from 0 to 1 of (x+2)(x-3)

\int_{0}^{1} (x + 2) (x - 3) d x

- \frac{37}{6}

Decimal Notation

- 6.16666 \dots

Solution steps

\int_{0}^{1} (x + 2) (x - 3) d x

Expand

(x + 2) (x - 3) : x^{2} - x - 6

= \int_{0}^{1} x^{2} - x - 6 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} 6 d x

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

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

\int_{0}^{1} 6 d x = 6

= \frac{1}{3} - \frac{1}{2} - 6

Simplify

\frac{1}{3} - \frac{1}{2} - 6 : - \frac{37}{6}

= - \frac{37}{6}

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What is the integral from 0 to 1 of (x+2)(x-3) ?
The integral from 0 to 1 of (x+2)(x-3) is -37/6