What is the integral of 0.02+0.12sqrt(x) ?

The integral of 0.02+0.12sqrt(x) is 0.02x+0.08x^{3/2}+C

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integral of 0.02+0.12sqrt(x)

\int 0.02 + 0.12 x d x

0.02 x + 0.08 x^{\frac{3}{2}} + C

Solution steps

\int 0.02 + 0.12 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 0.02 d x + \int 0.12 x d x

\int 0.02 d x = 0.02 x

\int 0.12 x d x = 0.08 x^{\frac{3}{2}}

= 0.02 x + 0.08 x^{\frac{3}{2}}

Add a constant to the solution

= 0.02 x + 0.08 x^{\frac{3}{2}} + C

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What is the integral of 0.02+0.12sqrt(x) ?
The integral of 0.02+0.12sqrt(x) is 0.02x+0.08x^{3/2}+C