What is the integral from 3 to 4 of (x-3)(4-x) ?

The integral from 3 to 4 of (x-3)(4-x) is 1/6

integral from 3 to 4 of (x-3)(4-x)

\int_{3}^{4} (x - 3) (4 - x) d x

\frac{1}{6}

Decimal

0.16666 \dots

Solution steps

\int_{3}^{4} (x - 3) (4 - x) d x

Expand

(x - 3) (4 - x) : 7 x - x^{2} - 12

= \int_{3}^{4} 7 x - x^{2} - 12 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_{3}^{4} 7 x d x - \int_{3}^{4} x^{2} d x - \int_{3}^{4} 12 d x

\int_{3}^{4} 7 x d x = \frac{49}{2}

\int_{3}^{4} x^{2} d x = \frac{37}{3}

\int_{3}^{4} 12 d x = 12

= \frac{49}{2} - \frac{37}{3} - 12

Simplify

\frac{49}{2} - \frac{37}{3} - 12 : \frac{1}{6}

= \frac{1}{6}

\int_{0}^{π} 2 - 2 cos (t) d t

\int_{0}^{2} \frac{x}{1 + 10 x ^{2}} d x

\int_{1}^{8} (\frac{1}{x}) d x

\int_{0}^{2} 4 - x d x

\int_{- 3}^{1} 8 d x

What is the integral from 3 to 4 of (x-3)(4-x) ?
The integral from 3 to 4 of (x-3)(4-x) is 1/6