求定积分从0到a[a-根号下(2ax-x^2)]/根号下(2a-x),请写详细点谢了 0到2a上 ∫ 根号下(2ax-x^2)dx
\u6c42\u5b9a\u79ef\u5206\u4ece0\u5230a[a-\u6839\u53f7\u4e0b(2ax-x^2)]/\u6839\u53f7\u4e0b(2a-x)\uff0c\u8bf7\u5199\u8be6\u7ec6\u70b9\u8c22\u4e86\u539f\u5f0f=\u222b(a-\u221a2ax-x^2)/\u221a2a-x
dx
\u79ef\u5206\u533a\u95f4(0\uff0ca)
=\u222b(a/\u221a2a-x)dx-\u221a2ax-x^2/\u221a2a-x
dx
\u79ef\u5206\u533a\u95f4(0\uff0ca)
=-a\u222b1/\u221a2a-x
d(2a-x)-
\u222b
\u221ax
dx
\u79ef\u5206\u533a\u95f4(0\uff0ca)
=-2a\u221a2a-x
-(2/3)x^(3/2)
\u79ef\u5206\u533a\u95f4(0\uff0ca)
=[2\u221a2-(8/3)]a^(3/2)
\u4ee5\u4e0a\u7b54\u6848\u4ec5\u4f9b\u53c2\u8003\uff0c\u5982\u6709\u7591\u95ee\u53ef\u7ee7\u7eed\u8ffd\u95ee!
\u222b [0,2a]\u221a\uff082ax-x^2\uff09dx
=\u222b [0,2a]\u221a[a^2-(a-x)^2]dx
=\u222b [\u03c0/2,-\u03c0/2]|a|*cost*(-a)costdt\uff08\u8bbea-x=asint\uff09
=a|a|\u222b [-\u03c0/2,\u03c0/2](1+cos2t)/2*dt
=a|a|(1/2*t+1/4*sin2t)| [\u03c0/2,-\u03c0/2]
=a|a|\u03c0/2
=∫(a/√2a-x)dx-√2ax-x^2/√2a-x dx 积分区间(0,a)
=-a∫1/√2a-x d(2a-x)- ∫ √x dx 积分区间(0,a)
=-2a√2a-x -(2/3)x^(3/2) 积分区间(0,a)
=[2√2-(8/3)]a^(3/2)
以上答案仅供参考,如有疑问可继续追问!
原式=∫(a-√2ax-x^2)/√2a-x dx 积分区间(0,a)
=∫(a/√2a-x)dx-√2ax-x^2/√2a-x dx 积分区间(0,a)
=-a∫1/√2a-x d(2a-x)- ∫ √x dx 积分区间(0,a)
=-2a√2a-x -(2/3)x^(3/2) 积分区间(0,a)
=[2√2-(8/3)]a^(3/2)
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