What is the integral of 14-20-3sqrt(16-2x) ?

The integral of 14-20-3sqrt(16-2x) is -6x+(16-2x)^{3/2}+C

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integral of 14-20-3sqrt(16-2x)

\int 14 - 20 - 3 16 - 2 x d x

- 6 x + (16 - 2 x)^{\frac{3}{2}} + C

Solution steps

\int 14 - 20 - 3 16 - 2 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 14 d x - \int 20 d x - \int 3 16 - 2 x d x

\int 14 d x = 14 x

\int 20 d x = 20 x

\int 3 16 - 2 x d x = - (16 - 2 x)^{\frac{3}{2}}

= 14 x - 20 x - (- (16 - 2 x)^{\frac{3}{2}})

Simplify

= - 6 x + (16 - 2 x)^{\frac{3}{2}}

Add a constant to the solution

= - 6 x + (16 - 2 x)^{\frac{3}{2}} + C

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What is the integral of 14-20-3sqrt(16-2x) ?
The integral of 14-20-3sqrt(16-2x) is -6x+(16-2x)^{3/2}+C