What is the integral from 4 to 8 of (4-x)(x-8) ?

The integral from 4 to 8 of (4-x)(x-8) is 32/3

integral from 4 to 8 of (4-x)(x-8)

\int_{4}^{8} (4 - x) (x - 8) d x

\frac{32}{3}

Decimal

10.66666 \dots

Solution steps

\int_{4}^{8} (4 - x) (x - 8) d x

Expand

(4 - x) (x - 8) : 12 x - x^{2} - 32

= \int_{4}^{8} 12 x - x^{2} - 32 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_{4}^{8} 12 x d x - \int_{4}^{8} x^{2} d x - \int_{4}^{8} 32 d x

\int_{4}^{8} 12 x d x = 288

\int_{4}^{8} x^{2} d x = \frac{448}{3}

\int_{4}^{8} 32 d x = 128

= 288 - \frac{448}{3} - 128

Simplify

288 - \frac{448}{3} - 128 : \frac{32}{3}

= \frac{32}{3}

\int_{0}^{12} \frac{10 x}{x ^{2} + 1} d x

\int_{0}^{1} \frac{1}{3} (x + y) d y

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

\int_{0}^{1} e^{x - e^{x}} d x

\int_{\frac{1}{2}}^{\frac{3}{4}} 2 x d x

What is the integral from 4 to 8 of (4-x)(x-8) ?
The integral from 4 to 8 of (4-x)(x-8) is 32/3