What is the integral from 0 to 1000 of 0.03x+0.1 ?

The integral from 0 to 1000 of 0.03x+0.1 is 15100

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integral from 0 to 1000 of 0.03x+0.1

\int_{0}^{1000} 0.03 x + 0.1 d x

15100

Solution steps

\int_{0}^{1000} 0.03 x + 0.1 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}^{1000} 0.03 x d x + \int_{0}^{1000} 0.1 d x

\int_{0}^{1000} 0.03 x d x = 15000

\int_{0}^{1000} 0.1 d x = 100

= 15000 + 100

Simplify

= 15100

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What is the integral from 0 to 1000 of 0.03x+0.1 ?
The integral from 0 to 1000 of 0.03x+0.1 is 15100