What is the integral of 32.5-12.5\sqrt[3]{x+0.5} ?

The integral of 32.5-12.5\sqrt[3]{x+0.5} is 32.5(x+0.5)-9.375(x+0.5)^{4/3}+C

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integral of 32.5-12.5\sqrt[3]{x+0.5}

\int 32.5 - 12.5 3 x + 0.5 d x

32.5 (x + 0.5) - 9.375 (x + 0.5)^{\frac{4}{3}} + C

Solution steps

\int 32.5 - 12.5 3 x + 0.5 d x

Apply u-substitution

: \int 32.5 - 12.5 3 u d u

= \int 32.5 - 12.5 3 u d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int 32.5 d u - \int 12.5 3 u d u

\int 32.5 d u = 32.5 u

\int 12.5 3 u d u = 9.375 u^{\frac{4}{3}}

= 32.5 u - 9.375 u^{\frac{4}{3}}

Substitute back

u = x + 0.5

= 32.5 (x + 0.5) - 9.375 (x + 0.5)^{\frac{4}{3}}

Add a constant to the solution

= 32.5 (x + 0.5) - 9.375 (x + 0.5)^{\frac{4}{3}} + C

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What is the integral of 32.5-12.5\sqrt[3]{x+0.5} ?
The integral of 32.5-12.5\sqrt[3]{x+0.5} is 32.5(x+0.5)-9.375(x+0.5)^{4/3}+C