What is the integral of (8-x^3)^2 ?

The integral of (8-x^3)^2 is 64x-4x^4+(x^7)/7+C

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integral of (8-x^3)^2

\int (8 - x^{3})^{2} d x

64 x - 4 x^{4} + \frac{x ^{7}}{7} + C

Solution steps

\int (8 - x^{3})^{2} d x

Expand

(8 - x^{3})^{2} : 64 - 16 x^{3} + x^{6}

= \int 64 - 16 x^{3} + x^{6} 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 64 d x - \int 16 x^{3} d x + \int x^{6} d x

\int 64 d x = 64 x

\int 16 x^{3} d x = 4 x^{4}

\int x^{6} d x = \frac{x ^{7}}{7}

= 64 x - 4 x^{4} + \frac{x ^{7}}{7}

Add a constant to the solution

= 64 x - 4 x^{4} + \frac{x ^{7}}{7} + C

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What is the integral of (8-x^3)^2 ?
The integral of (8-x^3)^2 is 64x-4x^4+(x^7)/7+C