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Popular >

integral of-0.02x+16

Solution

\int - 0.02 x + 16 d x

Solution

- 0.01 x^{2} + 16 x + C

Solution steps

\int - 0.02 x + 16 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 0.02 x d x + \int 16 d x

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

\int 16 d x = 16 x

= - 0.01 x^{2} + 16 x

Add a constant to the solution

= - 0.01 x^{2} + 16 x + C

Graph

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Frequently Asked Questions (FAQ)

What is the integral of-0.02x+16 ?
The integral of-0.02x+16 is -0.01x^2+16x+C