What is the integral of 200-0.02x ?

The integral of 200-0.02x is 200x-0.01x^2+C

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integral of 200-0.02x

\int 200 - 0.02 x d x

200 x - 0.01 x^{2} + C

Solution steps

\int 200 - 0.02 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

= \int 200 d x - \int 0.02 x d x

\int 200 d x = 200 x

\int 0.02 x d x = 0.01 x^{2}

= 200 x - 0.01 x^{2}

Add a constant to the solution

= 200 x - 0.01 x^{2} + C

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