What is the integral of 6(2-x)^5 ?

The integral of 6(2-x)^5 is 192x-240x^2+160x^3-60x^4+12x^5-x^6+C

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integral of 6(2-x)^5

integral

6 (2 - x)^{5}

192 x - 240 x^{2} + 160 x^{3} - 60 x^{4} + 12 x^{5} - x^{6} + C

Solution steps

\int 6 (2 - x)^{5} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 6 \cdot \int (2 - x)^{5} d x

Expand

(2 - x)^{5} : 32 - 80 x + 80 x^{2} - 40 x^{3} + 10 x^{4} - x^{5}

= 6 \cdot \int 32 - 80 x + 80 x^{2} - 40 x^{3} + 10 x^{4} - 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

= 6 (\int 32 d x - \int 80 x d x + \int 80 x^{2} d x - \int 40 x^{3} d x + \int 10 x^{4} d x - \int x^{5} d x)

\int 32 d x = 32 x

\int 80 x d x = 40 x^{2}

\int 80 x^{2} d x = \frac{80 x ^{3}}{3}

\int 40 x^{3} d x = 10 x^{4}

\int 10 x^{4} d x = 2 x^{5}

\int x^{5} d x = \frac{x ^{6}}{6}

= 6 (32 x - 40 x^{2} + \frac{80 x ^{3}}{3} - 10 x^{4} + 2 x^{5} - \frac{x ^{6}}{6})

Simplify

6 (32 x - 40 x^{2} + \frac{80 x ^{3}}{3} - 10 x^{4} + 2 x^{5} - \frac{x ^{6}}{6}) : 192 x - 240 x^{2} + 160 x^{3} - 60 x^{4} + 12 x^{5} - x^{6}

= 192 x - 240 x^{2} + 160 x^{3} - 60 x^{4} + 12 x^{5} - x^{6}

Add a constant to the solution

= 192 x - 240 x^{2} + 160 x^{3} - 60 x^{4} + 12 x^{5} - x^{6} + C

What is the integral of 6(2-x)^5 ?
The integral of 6(2-x)^5 is 192x-240x^2+160x^3-60x^4+12x^5-x^6+C