What is the integral from 1 to 2 of x^2(x-2)^{10} ?

The integral from 1 to 2 of x^2(x-2)^{10} is 46/429

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integral from 1 to 2 of x^2(x-2)^{10}

\int_{1}^{2} x^{2} (x - 2)^{10} d x

\frac{46}{429}

Decimal

0.10722 \dots

Solution steps

\int_{1}^{2} x^{2} (x - 2)^{10} d x

Expand

x^{2} (x - 2)^{10} : x^{12} - 20 x^{11} + 180 x^{10} - 960 x^{9} + 3360 x^{8} - 8064 x^{7} + 13440 x^{6} - 15360 x^{5} + 11520 x^{4} - 5120 x^{3} + 1024 x^{2}

= \int_{1}^{2} x^{12} - 20 x^{11} + 180 x^{10} - 960 x^{9} + 3360 x^{8} - 8064 x^{7} + 13440 x^{6} - 15360 x^{5} + 11520 x^{4} - 5120 x^{3} + 1024 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_{1}^{2} x^{12} d x - \int_{1}^{2} 20 x^{11} d x + \int_{1}^{2} 180 x^{10} d x - \int_{1}^{2} 960 x^{9} d x + \int_{1}^{2} 3360 x^{8} d x - \int_{1}^{2} 8064 x^{7} d x + \int_{1}^{2} 13440 x^{6} d x - \int_{1}^{2} 15360 x^{5} d x + \int_{1}^{2} 11520 x^{4} d x - \int_{1}^{2} 5120 x^{3} d x + \int_{1}^{2} 1024 x^{2} d x

\int_{1}^{2} x^{12} d x = \frac{8191}{13}

\int_{1}^{2} 20 x^{11} d x = 6825

\int_{1}^{2} 180 x^{10} d x = \frac{368460}{11}

\int_{1}^{2} 960 x^{9} d x = 98208

\int_{1}^{2} 3360 x^{8} d x = \frac{572320}{3}

\int_{1}^{2} 8064 x^{7} d x = 257040

\int_{1}^{2} 13440 x^{6} d x = 243840

\int_{1}^{2} 15360 x^{5} d x = 161280

\int_{1}^{2} 11520 x^{4} d x = 71424

\int_{1}^{2} 5120 x^{3} d x = 19200

\int_{1}^{2} 1024 x^{2} d x = \frac{7168}{3}

= \frac{8191}{13} - 6825 + \frac{368460}{11} - 98208 + \frac{572320}{3} - 257040 + 243840 - 161280 + 71424 - 19200 + \frac{7168}{3}

Simplify

\frac{8191}{13} - 6825 + \frac{368460}{11} - 98208 + \frac{572320}{3} - 257040 + 243840 - 161280 + 71424 - 19200 + \frac{7168}{3} : \frac{46}{429}

= \frac{46}{429}

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What is the integral from 1 to 2 of x^2(x-2)^{10} ?
The integral from 1 to 2 of x^2(x-2)^{10} is 46/429