What is the integral from 0 to 8 of (16x-128) ?

The integral from 0 to 8 of (16x-128) is -512

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integral from 0 to 8 of (16x-128)

\int_{0}^{8} (16 x - 128) d x

- 512

Solution steps

\int_{0}^{8} 16 x - 128 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}^{8} 16 x d x - \int_{0}^{8} 128 d x

\int_{0}^{8} 16 x d x = 512

\int_{0}^{8} 128 d x = 1024

= 512 - 1024

Simplify

= - 512

What is the integral from 0 to 8 of (16x-128) ?
The integral from 0 to 8 of (16x-128) is -512