What is the integral of x^2(1-x)^5 ?

The integral of x^2(1-x)^5 is (x^3)/3-(5x^4)/4+2x^5-(5x^6)/3+(5x^7)/7-(x^8)/8+C

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integral of x^2(1-x)^5

\int x^{2} (1 - x)^{5} d x

\frac{x ^{3}}{3} - \frac{5 x ^{4}}{4} + 2 x^{5} - \frac{5 x ^{6}}{3} + \frac{5 x ^{7}}{7} - \frac{x ^{8}}{8} + C

Solution steps

\int x^{2} (1 - x)^{5} d x

Expand

x^{2} (1 - x)^{5} : x^{2} - 5 x^{3} + 10 x^{4} - 10 x^{5} + 5 x^{6} - x^{7}

= \int x^{2} - 5 x^{3} + 10 x^{4} - 10 x^{5} + 5 x^{6} - 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 x^{2} d x - \int 5 x^{3} d x + \int 10 x^{4} d x - \int 10 x^{5} d x + \int 5 x^{6} d x - \int x^{7} d x

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

\int 5 x^{3} d x = \frac{5 x ^{4}}{4}

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

\int 10 x^{5} d x = \frac{5 x ^{6}}{3}

\int 5 x^{6} d x = \frac{5 x ^{7}}{7}

\int x^{7} d x = \frac{x ^{8}}{8}

= \frac{x ^{3}}{3} - \frac{5 x ^{4}}{4} + 2 x^{5} - \frac{5 x ^{6}}{3} + \frac{5 x ^{7}}{7} - \frac{x ^{8}}{8}

Add a constant to the solution

= \frac{x ^{3}}{3} - \frac{5 x ^{4}}{4} + 2 x^{5} - \frac{5 x ^{6}}{3} + \frac{5 x ^{7}}{7} - \frac{x ^{8}}{8} + C

\int \frac{x ^{3}}{x ^{2} + 2 x + 5} d x

\frac{d}{d t} (- e^{- t^{2}})

y = x^{2}, y = 4, x = 0

\int_{0}^{7} π (2 5 y)^{2} d y

\frac{d y}{d x} = sin (x - y)

What is the integral of x^2(1-x)^5 ?
The integral of x^2(1-x)^5 is (x^3)/3-(5x^4)/4+2x^5-(5x^6)/3+(5x^7)/7-(x^8)/8+C