What is the integral from 0 to 1 of (r+3)(r-4) ?

The integral from 0 to 1 of (r+3)(r-4) is -73/6

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integral from 0 to 1 of (r+3)(r-4)

\int_{0}^{1} (r + 3) (r - 4) d r

- \frac{73}{6}

Decimal

- 12.16666 \dots

Solution steps

\int_{0}^{1} (r + 3) (r - 4) d r

Expand

(r + 3) (r - 4) : r^{2} - r - 12

= \int_{0}^{1} r^{2} - r - 12 d r

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} r^{2} d r - \int_{0}^{1} r d r - \int_{0}^{1} 12 d r

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

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

\int_{0}^{1} 12 d r = 12

= \frac{1}{3} - \frac{1}{2} - 12

Simplify

\frac{1}{3} - \frac{1}{2} - 12 : - \frac{73}{6}

= - \frac{73}{6}

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What is the integral from 0 to 1 of (r+3)(r-4) ?
The integral from 0 to 1 of (r+3)(r-4) is -73/6