What is the integral of (2-2x)^7 ?

The integral of (2-2x)^7 is 128x-448x^2+896x^3-1120x^4+896x^5-448x^6+128x^7-16x^8+C

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integral of (2-2x)^7

\int (2 - 2 x)^{7} d x

128 x - 448 x^{2} + 896 x^{3} - 1120 x^{4} + 896 x^{5} - 448 x^{6} + 128 x^{7} - 16 x^{8} + C

Solution steps

\int (2 - 2 x)^{7} d x

Expand

(2 - 2 x)^{7} : 128 - 896 x + 2688 x^{2} - 4480 x^{3} + 4480 x^{4} - 2688 x^{5} + 896 x^{6} - 128 x^{7}

= \int 128 - 896 x + 2688 x^{2} - 4480 x^{3} + 4480 x^{4} - 2688 x^{5} + 896 x^{6} - 128 x^{7} 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 128 d x - \int 896 x d x + \int 2688 x^{2} d x - \int 4480 x^{3} d x + \int 4480 x^{4} d x - \int 2688 x^{5} d x + \int 896 x^{6} d x - \int 128 x^{7} d x

\int 128 d x = 128 x

\int 896 x d x = 448 x^{2}

\int 2688 x^{2} d x = 896 x^{3}

\int 4480 x^{3} d x = 1120 x^{4}

\int 4480 x^{4} d x = 896 x^{5}

\int 2688 x^{5} d x = 448 x^{6}

\int 896 x^{6} d x = 128 x^{7}

\int 128 x^{7} d x = 16 x^{8}

= 128 x - 448 x^{2} + 896 x^{3} - 1120 x^{4} + 896 x^{5} - 448 x^{6} + 128 x^{7} - 16 x^{8}

Add a constant to the solution

= 128 x - 448 x^{2} + 896 x^{3} - 1120 x^{4} + 896 x^{5} - 448 x^{6} + 128 x^{7} - 16 x^{8} + C

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What is the integral of (2-2x)^7 ?
The integral of (2-2x)^7 is 128x-448x^2+896x^3-1120x^4+896x^5-448x^6+128x^7-16x^8+C