What is the integral from 0 to 16 of 0.8(16-x) ?

The integral from 0 to 16 of 0.8(16-x) is 102.4

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integral from 0 to 16 of 0.8(16-x)

\int_{0}^{16} 0.8 (16 - x) d x

102.4

Solution steps

\int_{0}^{16} 0.8 (16 - x) d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 0.8 \cdot \int_{0}^{16} 16 - x 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

= 0.8 (\int_{0}^{16} 16 d x - \int_{0}^{16} x d x)

\int_{0}^{16} 16 d x = 256

\int_{0}^{16} x d x = 128

= 0.8 (256 - 128)

Simplify

= 102.4

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What is the integral from 0 to 16 of 0.8(16-x) ?
The integral from 0 to 16 of 0.8(16-x) is 102.4