What is the integral from 0 to 1 of x^2(x+1)^6 ?

The integral from 0 to 1 of x^2(x+1)^6 is 2815/252

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integral from 0 to 1 of x^2(x+1)^6

\int_{0}^{1} x^{2} (x + 1)^{6} d x

\frac{2815}{252}

Decimal

11.17063 \dots

Solution steps

\int_{0}^{1} x^{2} (x + 1)^{6} d x

Expand

x^{2} (x + 1)^{6} : x^{8} + 6 x^{7} + 15 x^{6} + 20 x^{5} + 15 x^{4} + 6 x^{3} + x^{2}

= \int_{0}^{1} x^{8} + 6 x^{7} + 15 x^{6} + 20 x^{5} + 15 x^{4} + 6 x^{3} + x^{2} 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_{0}^{1} x^{8} d x + \int_{0}^{1} 6 x^{7} d x + \int_{0}^{1} 15 x^{6} d x + \int_{0}^{1} 20 x^{5} d x + \int_{0}^{1} 15 x^{4} d x + \int_{0}^{1} 6 x^{3} d x + \int_{0}^{1} x^{2} d x

\int_{0}^{1} x^{8} d x = \frac{1}{9}

\int_{0}^{1} 6 x^{7} d x = \frac{3}{4}

\int_{0}^{1} 15 x^{6} d x = \frac{15}{7}

\int_{0}^{1} 20 x^{5} d x = \frac{10}{3}

\int_{0}^{1} 15 x^{4} d x = 3

\int_{0}^{1} 6 x^{3} d x = \frac{3}{2}

\int_{0}^{1} x^{2} d x = \frac{1}{3}

= \frac{1}{9} + \frac{3}{4} + \frac{15}{7} + \frac{10}{3} + 3 + \frac{3}{2} + \frac{1}{3}

Simplify

\frac{1}{9} + \frac{3}{4} + \frac{15}{7} + \frac{10}{3} + 3 + \frac{3}{2} + \frac{1}{3} : \frac{2815}{252}

= \frac{2815}{252}

\frac{6}{s ( s + 3 )}

\frac{d}{d x} (\frac{1}{1 + 5 x})

\frac{d}{d x} (a sin^{3} (x))

x \to 0 - lim (\frac{14 x ^{3} + 15 x}{x})

ln (\frac{1}{3} (2 + e^{- 3 x}))

What is the integral from 0 to 1 of x^2(x+1)^6 ?
The integral from 0 to 1 of x^2(x+1)^6 is 2815/252