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

The integral of (2-x)^5 is 32x-40x^2+(80x^3)/3-10x^4+2x^5-(x^6)/6+C

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integral of (2-x)^5

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

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

Solution steps

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

Expand

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

= \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

= \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}

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

Add a constant to the solution

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

\int_{0}^{9} x 81 - x^{2} d x

\frac{d}{d x} (e^{2 x} + 2)

\frac{s ^{2}}{s + 1}

x y^{^{'}} = y + 2 x^{2} sin (x), y (π) = 0

n = 1 \sum \infty \frac{e ^{n}}{n ^{2}}

What is the integral of (2-x)^5 ?
The integral of (2-x)^5 is 32x-40x^2+(80x^3)/3-10x^4+2x^5-(x^6)/6+C