What is the integral of (4-4x)^5 ?

The integral of (4-4x)^5 is 1024x-2560x^2+(10240x^3)/3-2560x^4+1024x^5-(512x^6)/3+C

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integral of (4-4x)^5

\int (4 - 4 x)^{5} d x

1024 x - 2560 x^{2} + \frac{10240 x ^{3}}{3} - 2560 x^{4} + 1024 x^{5} - \frac{512 x ^{6}}{3} + C

Solution steps

\int (4 - 4 x)^{5} d x

Expand

(4 - 4 x)^{5} : 1024 - 5120 x + 10240 x^{2} - 10240 x^{3} + 5120 x^{4} - 1024 x^{5}

= \int 1024 - 5120 x + 10240 x^{2} - 10240 x^{3} + 5120 x^{4} - 1024 x^{5} 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 1024 d x - \int 5120 x d x + \int 10240 x^{2} d x - \int 10240 x^{3} d x + \int 5120 x^{4} d x - \int 1024 x^{5} d x

\int 1024 d x = 1024 x

\int 5120 x d x = 2560 x^{2}

\int 10240 x^{2} d x = \frac{10240 x ^{3}}{3}

\int 10240 x^{3} d x = 2560 x^{4}

\int 5120 x^{4} d x = 1024 x^{5}

\int 1024 x^{5} d x = \frac{512 x ^{6}}{3}

= 1024 x - 2560 x^{2} + \frac{10240 x ^{3}}{3} - 2560 x^{4} + 1024 x^{5} - \frac{512 x ^{6}}{3}

Add a constant to the solution

= 1024 x - 2560 x^{2} + \frac{10240 x ^{3}}{3} - 2560 x^{4} + 1024 x^{5} - \frac{512 x ^{6}}{3} + C

What is the integral of (4-4x)^5 ?
The integral of (4-4x)^5 is 1024x-2560x^2+(10240x^3)/3-2560x^4+1024x^5-(512x^6)/3+C