第1个回答 2013-05-28
计算定积分 ∫(x+3)/根号(2x+1)dx,上限4,下限0
解:先计算不定积分,不考虑积分后的待定常数项C
∫[(x+3)/sqrt(2x+1)]dx
=∫{[(2x+1)/2+3/2]/sqrt(2x+1)}dx
=∫{[(2x+1)/2]/sqrt(2x+1)}dx+∫[(3/2)/sqrt(2x+1)]dx
=1/2∫(2x+1)^(1/2)dx+3/2∫dx/sqrt(2x+1)
=1/2*1/2∫(2x+1)^(1/2)d(2x+1)+3/2*1/2∫(2x+1)^(-1/2)d(2x+1)
=1/4*1/(1+1/2)(2x+1)^(1+1/2)+3/4*1/(1-1/2)(2x+1)^(1-1-2)
=1/4*(2/3)(2x+1)^(3/2)+3/4*2(2x+1)^(1/2)
=2/3*(2x+1)^(3/2)+3/2*(2x+1)^(1/2), 然后代入积分上下限:上限4,下限0
=2/3*(2*4+1)^(3/2)+3/2*(2*4+1)^(1/2)-2/3*(2*0+1)^(3/2)-3/2*(2*0+1)^(1/2)
=2/3*27+3/2*3-2/3*1-3/2*1
=18+9/2-2/3-3/2
=21-2/3
=20+(1/3)
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