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

解答

\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}

求解步骤

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

乘开

(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

使用积分加法定则:

\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}

\int_{- 1}^{0} \frac{2 t}{( 2 + t ^{2} ) ^{3}} d t

\int_{0}^{1} 0.5 d x

\int_{0}^{2} x^{2} - x - 2 d x

\int_{1}^{x^{2}} 3 + x^{2} d x

\int_{0}^{l n (a)} 6 e^{3 x} d x