What is the integral of ((8-x)(8+x)) ?

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

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integral of ((8-x)(8+x))

\int ((8 - x) (8 + x)) d x

64 x - \frac{x ^{3}}{3} + C

Solution steps

\int (8 - x) (8 + x) d x

Expand

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

= \int 64 - x^{2} 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 x^{2} d x

\int 64 d x = 64 x

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

= 64 x - \frac{x ^{3}}{3}

Add a constant to the solution

= 64 x - \frac{x ^{3}}{3} + C

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