What is the integral from y^3 to y of (1-y^2)^8 ?

The integral from y^3 to y of (1-y^2)^8 is

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integral from y^3 to y of (1-y^2)^8

\int_{y^{3}}^{y} (1 - y^{2})^{8} d y

y - y^{3} - \frac{8 ( y ^{3} - y ^{9} )}{3} + \frac{28 ( y ^{5} - y ^{15} )}{5} - 8 (- y^{21} + y^{7}) + \frac{70 ( y ^{9} - y ^{27} )}{9} - \frac{56 ( y ^{11} - y ^{33} )}{11} + \frac{28 ( y ^{13} - y ^{39} )}{13} - \frac{8 ( y ^{15} - y ^{45} )}{15} + \frac{y ^{17} - y ^{51}}{17}

Solution steps

\int_{y^{3}}^{y} (1 - y^{2})^{8} d y

Expand

(1 - y^{2})^{8} : 1 - 8 y^{2} + 28 y^{4} - 56 y^{6} + 70 y^{8} - 56 y^{10} + 28 y^{12} - 8 y^{14} + y^{16}

= \int_{y^{3}}^{y} 1 - 8 y^{2} + 28 y^{4} - 56 y^{6} + 70 y^{8} - 56 y^{10} + 28 y^{12} - 8 y^{14} + y^{16} d y

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{y^{3}}^{y} 1 d y - \int_{y^{3}}^{y} 8 y^{2} d y + \int_{y^{3}}^{y} 28 y^{4} d y - \int_{y^{3}}^{y} 56 y^{6} d y + \int_{y^{3}}^{y} 70 y^{8} d y - \int_{y^{3}}^{y} 56 y^{10} d y + \int_{y^{3}}^{y} 28 y^{12} d y - \int_{y^{3}}^{y} 8 y^{14} d y + \int_{y^{3}}^{y} y^{16} d y

\int_{y^{3}}^{y} 1 d y = y - y^{3}

\int_{y^{3}}^{y} 8 y^{2} d y = \frac{8 ( y ^{3} - y ^{9} )}{3}

\int_{y^{3}}^{y} 28 y^{4} d y = \frac{28 ( y ^{5} - y ^{15} )}{5}

\int_{y^{3}}^{y} 56 y^{6} d y = 8 (- y^{21} + y^{7})

\int_{y^{3}}^{y} 70 y^{8} d y = \frac{70 ( y ^{9} - y ^{27} )}{9}

\int_{y^{3}}^{y} 56 y^{10} d y = \frac{56 ( y ^{11} - y ^{33} )}{11}

\int_{y^{3}}^{y} 28 y^{12} d y = \frac{28 ( y ^{13} - y ^{39} )}{13}

\int_{y^{3}}^{y} 8 y^{14} d y = \frac{8 ( y ^{15} - y ^{45} )}{15}

\int_{y^{3}}^{y} y^{16} d y = \frac{y ^{17} - y ^{51}}{17}

= y - y^{3} - \frac{8 ( y ^{3} - y ^{9} )}{3} + \frac{28 ( y ^{5} - y ^{15} )}{5} - 8 (- y^{21} + y^{7}) + \frac{70 ( y ^{9} - y ^{27} )}{9} - \frac{56 ( y ^{11} - y ^{33} )}{11} + \frac{28 ( y ^{13} - y ^{39} )}{13} - \frac{8 ( y ^{15} - y ^{45} )}{15} + \frac{y ^{17} - y ^{51}}{17}

What is the integral from y^3 to y of (1-y^2)^8 ?
The integral from y^3 to y of (1-y^2)^8 is