求定积分√(2-x^2),上限为√2,下限为0设t=√(2-x^2),x=√(2-t^2),dx=[(2-x^2)]^(-1/2)dt;当x=0,t=√2,当x=√2,t=0;∫_0^√2[√(2-x^2)]dx=∫_√2^0{t[(2-x^2)]^(-1/2)}dt=(-1/2)∫_√2^0{[(2-x^2)]^(-1/2)}d(2-x^2)=(-1/2)[2-x^2](上为0,
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求定积分√(2-x^2),上限为√2,下限为0设t=√(2-x^2),x=√(2-t^2),dx=[(2-x^2)]^(-1/2)dt;当x=0,t=√2,当x=√2,t=0;∫_0^√2[√(2-x^2)]dx=∫_√2^0{t[(2-x^2)]^(-1/2)}dt=(-1/2)∫_√2^0{[(2-x^2)]^(-1/2)}d(2-x^2)=(-1/2)[2-x^2](上为0,
求定积分√(2-x^2),上限为√2,下限为0
设t=√(2-x^2),x=√(2-t^2),dx=[(2-x^2)]^(-1/2)dt;当x=0,t=√2,当x=√2,t=0;
∫_0^√2[√(2-x^2)]dx=∫_√2^0{t[(2-x^2)]^(-1/2)}dt
=(-1/2)∫_√2^0{[(2-x^2)]^(-1/2)}d(2-x^2)
=(-1/2)[2-x^2](上为0,下为√2)
=-√2
(∫_0^√2表示定积分上限为√2,下限为0∫)
注:这是我做出的结果,但标准答案为∏/2,请找出我哪个步骤错了,并提示思路.
求定积分√(2-x^2),上限为√2,下限为0设t=√(2-x^2),x=√(2-t^2),dx=[(2-x^2)]^(-1/2)dt;当x=0,t=√2,当x=√2,t=0;∫_0^√2[√(2-x^2)]dx=∫_√2^0{t[(2-x^2)]^(-1/2)}dt=(-1/2)∫_√2^0{[(2-x^2)]^(-1/2)}d(2-x^2)=(-1/2)[2-x^2](上为0,
你的错误在:“dx=[(2-x^2)]^(-1/2)dt”!而且,中间的字母变换也搞
混淆了.
正确的是:“dx=-tdt/√(2-t²)”!
你这种思路完全错误了,反而把原定积分变换复杂了.正切的解法如下:
设x=√2sint,则dx=√2costdt.(说明:∫(a,b)表示从a到b积分)
∴原定积分=∫(0,π/2)[2cos²t]dt
=∫(0,π/2)[1+cos(2t)]dt
=[t+1/2sin(2t)]|(0,π/2)
=π/2+1/2sinπ-0-1/2sin0
=π/2.
应令t=(2-x^2)^1/2→x=(2-t^2)^1/2→dx=-t^2(2-t^2)^-1/2dt(复合求导令u=2-t^2则有x=u^1/2→dx=1/2(u)^-1/2du而du=-2tdt→dx/dt=dx/du*du/dt代入得
∏/2 +1