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

The integral from 0 to 1 of x(1+x)^6 is 769/56

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

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

\frac{769}{56}

Decimal

13.73214 \dots

Solution steps

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

Expand

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

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

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

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

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

\int_{0}^{1} 20 x^{4} d x = 4

\int_{0}^{1} 15 x^{5} d x = \frac{5}{2}

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

\int_{0}^{1} x^{7} d x = \frac{1}{8}

= \frac{1}{2} + 2 + \frac{15}{4} + 4 + \frac{5}{2} + \frac{6}{7} + \frac{1}{8}

Simplify

\frac{1}{2} + 2 + \frac{15}{4} + 4 + \frac{5}{2} + \frac{6}{7} + \frac{1}{8} : \frac{769}{56}

= \frac{769}{56}

\int_{0}^{1} 6 x^{3} d x

\int_{0}^{1} \frac{9}{1 + x ^{2}} d x

\int_{0}^{2} (x^{3} - x^{2} - 2 x) d x

\int_{0}^{3} 10 x d x

\int_{\frac{π}{2}}^{π} 1

What is the integral from 0 to 1 of x(1+x)^6 ?
The integral from 0 to 1 of x(1+x)^6 is 769/56