標題:
此文章來自奇摩知識+如有不便請留言告知
pure maths Q?
發問:
pure maths Q P(m)=∫x^m * √(2ax-x^2) dx prove that P(m)= - {[x^(m+1/2) * (2a-x)^(3/2)] / (m+2)} + {[(2am+a) / (m+2)] * P(m-1) } ^= power 更新: my e-mail: tongqu2@yahoo.com.hk
最佳解答:
P(m) = ∫x^m √(2ax-x^2) dx = -∫x^(m+1/2) √(2a-x) d(2a-x) = -(2/3)∫x^(m+1/2) d(2a-x)^(3/2) = (2/3)[ -x^(m+1/2) (2a-x)^(3/2) + ∫(2a-x)^(3/2) d(x^(m+1/2)) ] = (2/3)[ -x^(m+1/2) (2a-x)^(3/2) + (m+1/2)∫x^(m-1/2) (2a-x)^(3/2) dx ] = (2/3)[ -x^(m+1/2) (2a-x)^(3/2) + (m+1/2)∫x^(m-1/2) (2a-x)^(1/2) (2a-x) dx ] = (2/3)[ -x^(m+1/2) (2a-x)^(3/2) + (m+1/2)(2a)∫x^(m-1/2) (2a-x)^(1/2) dx - (m+1/2)∫x^(m+1/2) (2a-x)^(1/2) dx ] = (2/3)[ -x^(m+1/2) (2a-x)^(3/2) + 2a(2m+1)/2∫x^(m-1) √(2ax-x^2) dx - (m+1/2)∫x^m √(2ax-x^2) dx ] = (2/3)[ -x^(m+1/2) (2a-x)^(3/2) + (2am+a)P(m-1) - (m+1/2)P(m) ] Therefore (3/2)P(m) = -x^(m+1/2) (2a-x)^(3/2) + (2am+a)P(m-1) - (m+1/2)P(m) (m+2)P(m) = -x^(m+1/2) (2a-x)^(3/2) + (2am+a)P(m-1) P(m) = - {[x^(m+1/2) * (2a-x)^(3/2)] / (m+2)} + {[(2am+a) / (m+2)] * P(m-1) }
其他解答:
有冇email or msn,我send 比你睇
留言列表
