What is the integral from 0 to 1 of 2x(3x-2)^6 ?

The integral from 0 to 1 of 2x(3x-2)^6 is 31/28

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integral from 0 to 1 of 2x(3x-2)^6

\int_{0}^{1} 2 x (3 x - 2)^{6} d x

\frac{31}{28}

Decimal

1.10714 \dots

Solution steps

\int_{0}^{1} 2 x (3 x - 2)^{6} d x

Take the constant out:

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

= 2 \cdot \int_{0}^{1} x (3 x - 2)^{6} d x

Expand

x (3 x - 2)^{6} : 729 x^{7} - 2916 x^{6} + 4860 x^{5} - 4320 x^{4} + 2160 x^{3} - 576 x^{2} + 64 x

= 2 \cdot \int_{0}^{1} 729 x^{7} - 2916 x^{6} + 4860 x^{5} - 4320 x^{4} + 2160 x^{3} - 576 x^{2} + 64 x 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

= 2 (\int_{0}^{1} 729 x^{7} d x - \int_{0}^{1} 2916 x^{6} d x + \int_{0}^{1} 4860 x^{5} d x - \int_{0}^{1} 4320 x^{4} d x + \int_{0}^{1} 2160 x^{3} d x - \int_{0}^{1} 576 x^{2} d x + \int_{0}^{1} 64 x d x)

\int_{0}^{1} 729 x^{7} d x = \frac{729}{8}

\int_{0}^{1} 2916 x^{6} d x = \frac{2916}{7}

\int_{0}^{1} 4860 x^{5} d x = 810

\int_{0}^{1} 4320 x^{4} d x = 864

\int_{0}^{1} 2160 x^{3} d x = 540

\int_{0}^{1} 576 x^{2} d x = 192

\int_{0}^{1} 64 x d x = 32

= 2 (\frac{729}{8} - \frac{2916}{7} + 810 - 864 + 540 - 192 + 32)

Simplify

2 (\frac{729}{8} - \frac{2916}{7} + 810 - 864 + 540 - 192 + 32) : \frac{31}{28}

= \frac{31}{28}

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What is the integral from 0 to 1 of 2x(3x-2)^6 ?
The integral from 0 to 1 of 2x(3x-2)^6 is 31/28