What is the integral from 0 to 1 of (u+7)(u-8) ?

The integral from 0 to 1 of (u+7)(u-8) is -337/6

integral from 0 to 1 of (u+7)(u-8)

\int_{0}^{1} (u + 7) (u - 8) d u

- \frac{337}{6}

Decimal

- 56.16666 \dots

Solution steps

\int_{0}^{1} (u + 7) (u - 8) d u

Expand

(u + 7) (u - 8) : u^{2} - u - 56

= \int_{0}^{1} u^{2} - u - 56 d u

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} u^{2} d u - \int_{0}^{1} u d u - \int_{0}^{1} 56 d u

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

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

\int_{0}^{1} 56 d u = 56

= \frac{1}{3} - \frac{1}{2} - 56

Simplify

\frac{1}{3} - \frac{1}{2} - 56 : - \frac{337}{6}

= - \frac{337}{6}

\int 2 e^{4 x} d x

\frac{5}{3 ( s + 3 ) ^{2} + 27}

x \to - 1 lim (\frac{x}{x - 1})

f (x) = - x^{2} + 6 x - 8, 4, - 2

\int \frac{1}{6 x + 1} d x

What is the integral from 0 to 1 of (u+7)(u-8) ?
The integral from 0 to 1 of (u+7)(u-8) is -337/6